3.02h Motion under gravity: vector form

A particle is projected horizontally with a speed of 6 m s\(^{-1}\) from a point 10 m above horizontal ground. The particle moves freely under gravity. Calculate the speed and direction of motion of the particle at the instant it hits the ground. [6]

A particle is projected with speed 49 m s\(^{-1}\) at an angle of elevation \(\theta\) from a point \(O\) on a horizontal plane, and moves freely under gravity. The horizontal and upward vertical displacements of the particle from \(O\) at time \(t\) seconds after projection are \(x\) m and \(y\) m respectively.

1. Express \(x\) and \(y\) in terms of \(\theta\) and \(t\), and hence show that $$y = x \tan \theta - \frac{x^2(1 + \tan^2 \theta)}{490}.$$ [4]

\includegraphics{figure_8} The particle passes through the point where \(x = 70\) and \(y = 30\). The two possible values of \(\theta\) are \(\theta_1\) and \(\theta_2\), and the corresponding points where the particle returns to the plane are \(A_1\) and \(A_2\) respectively (see diagram).

1. Find \(\theta_1\) and \(\theta_2\). [4]
2. Calculate the distance between \(A_1\) and \(A_2\). [5]

A particle is projected with speed \(u\) ms\(^{-1}\) at an angle of \(\theta\) above the horizontal from a point \(O\). At time \(t\) s after projection, the horizontal and vertically upwards displacements of the particle from \(O\) are \(x\) m and \(y\) m respectively.

1. Express \(x\) and \(y\) in terms of \(t\) and \(\theta\) and hence obtain the equation of trajectory $$y = x \tan \theta - \frac{gx^2 \sec^2 \theta}{2u^2}.$$ [4]

In a shot put competition, a shot is thrown from a height of 2.1 m above horizontal ground. It has initial velocity of 14 ms\(^{-1}\) at an angle of \(\theta\) above the horizontal. The shot travels a horizontal distance of 22 m before hitting the ground.

1. Show that \(12.1 \tan^2 \theta - 22 \tan \theta + 10 = 0\), and find the value of \(\theta\). [5]
2. Find the time of flight of the shot. [2]

A particle is projected horizontally with a speed of \(7 \text{ ms}^{-1}\) from a point 10 m above horizontal ground. The particle moves freely under gravity. Calculate the speed and direction of motion of the particle at the instant it hits the ground. [6]

A small ball of mass 0.2 kg is projected with speed \(11 \text{ ms}^{-1}\) up a line of greatest slope of a roof from a point \(A\) at the bottom of the roof. The ball remains in contact with the roof and moves up the line of greatest slope to the top of the roof at \(B\). The roof is rough and the coefficient of friction is \(\frac{1}{4}\). The distance \(AB\) is 5 m and \(AB\) is inclined at \(30°\) to the horizontal (see diagram).

1. Show that the speed of the ball when it reaches \(B\) is \(5.44 \text{ ms}^{-1}\), correct to 2 decimal places. [6]

The ball leaves the roof at \(B\) and moves freely under gravity. The point \(C\) is at the lower edge of the roof. The distance \(BC\) is 5 m and \(BC\) is inclined at \(30°\) to the horizontal.

1. Determine whether or not the ball hits the roof between \(B\) and \(C\). [7]

A particle \(P\) is projected with speed \(32 \text{ m s}^{-1}\) at an angle of elevation \(\alpha\), where \(\sin \alpha = \frac{3}{4}\), from a point \(A\) on horizontal ground. At the same instant a particle \(Q\) is projected with speed \(20 \text{ m s}^{-1}\) at an angle of elevation \(\beta\), where \(\sin \beta = \frac{24}{25}\), from a point \(B\) on the same horizontal ground. The particles move freely under gravity in the same vertical plane and collide with each other at the point \(C\) at the instant when they are travelling horizontally (see diagram).

1. Calculate the height of \(C\) above the ground and the distance \(AB\). [4]

Immediately after the collision \(P\) falls vertically. \(P\) hits the ground and rebounds vertically upwards, coming to instantaneous rest at a height 5 m above the ground.

1. Given that the mass of \(P\) is 3 kg, find the magnitude and direction of the impulse exerted on \(P\) by the ground. [4]

The coefficient of restitution between the two particles is \(\frac{1}{2}\).

1. Find the distance of \(Q\) from \(C\) at the instant when \(Q\) is travelling in a direction of \(25°\) below the horizontal. [9]

Some tiles on a roof are being replaced. Each tile has a mass of 2 kg and the coefficient of friction between it and the existing roof is 0.75. The roof is at \(30°\) to the horizontal and the bottom of the roof is 6 m above horizontal ground, as shown in Fig. 4. \includegraphics{figure_4}

1. Calculate the limiting frictional force between a tile and the roof. A tile is placed on the roof. Does it slide? (Your answer should be supported by a calculation.) [5]
2. The tiles are raised 6 m from the ground, the only work done being against gravity. They are then slid 4 m up the roof and placed at the point A shown in Fig. 4.
   1. Show that each tile gains 156.8 J of gravitational potential energy. [3]
   2. Calculate the work done against friction per tile. [2]
   3. What average power is required to raise 10 tiles per minute from the ground to A? [2]
3. A tile is kicked from A directly down the roof. When the tile is at B, \(x\) m from the edge of the roof, its speed is \(4 \text{ m s}^{-1}\). It subsequently hits the ground travelling at \(9 \text{ m s}^{-1}\). In the motion of the tile from B to the ground, the work done against sliding and other resistances is 90 J. Use an energy method to find \(x\). [5]

A ball is projected from a point \(O\) on horizontal ground with speed \(14 \text{ m s}^{-1}\) at an angle of elevation \(30°\) above the horizontal. The ball travels in a vertical plane through the point \(O\) and hits a point \(Q\) on a plane which is inclined at \(45°\) to the horizontal. The point \(O\) is \(6\) metres from \(P\), the foot of the inclined plane, as shown in the diagram. The points \(O\), \(P\) and \(Q\) lie in the same vertical plane. The line \(PQ\) is a line of greatest slope of the inclined plane. \includegraphics{figure_3}

1. During its flight, the horizontal and upward vertical distances of the ball from \(O\) are \(x\) metres and \(y\) metres respectively. Show that \(x\) and \(y\) satisfy the equation $$y = x\frac{\sqrt{3}}{3} - \frac{x^2}{30}$$ Use \(\cos 30° = \frac{\sqrt{3}}{2}\) and \(\tan 30° = \frac{\sqrt{3}}{3}\). [5 marks]
2. Find the distance \(PQ\). [7 marks]

A ball is projected from a point \(O\) above a smooth plane which is inclined at an angle of \(20°\) to the horizontal. The point \(O\) is at a perpendicular distance of \(1\) m from the inclined plane. The ball is projected with velocity \(22 \text{ m s}^{-1}\) at an angle of \(70°\) above the horizontal. The motion of the ball is in a vertical plane containing a line of greatest slope of the inclined plane. The ball strikes the inclined plane for the first time at a point \(A\). \includegraphics{figure_5}

1. 1. Find the time taken by the ball to travel from \(O\) to \(A\). [4 marks]
   2. Find the components of the velocity of the ball, parallel and perpendicular to the inclined plane, as it strikes the plane at \(A\). [4 marks]
2. After striking \(A\), the ball rebounds and strikes the plane for a second time at a point further up than \(A\). The coefficient of restitution between the ball and the inclined plane is \(e\). Show that \(e < k\), where \(k\) is a constant to be determined. [4 marks]

A ball of mass \(m\) is thrown vertically upwards from the ground with an initial speed \(u\). When the speed of the ball is \(v\), the magnitude of the air resistance is \(mkv\), where \(k\) is a positive constant. By modelling the ball as a particle, find, in terms of \(u\), \(k\) and \(g\), the time taken for the ball to reach its greatest height. [8]

A small pebble of mass \(m\) is placed in a viscous liquid and sinks vertically from rest through the liquid. When the speed of the pebble is \(v\) the magnitude of the resistance due to the liquid is modelled as \(mkv^2\), where \(k\) is a positive constant. Find the speed of the pebble after it has fallen a distance \(D\) through the liquid. [11]

\includegraphics{figure_2} Boat \(A\) is travelling with constant speed 7.9 m s\(^{-1}\) on a course with bearing 035°. Boat \(B\) is travelling with constant speed 10.5 m s\(^{-1}\) on a course with bearing 330°. At one instant, the boats are 1500 m apart with \(B\) on a bearing of 125° from \(A\) (see diagram).

1. Find the magnitude and the bearing of the velocity of \(B\) relative to \(A\). [5]
2. Find the shortest distance between \(A\) and \(B\) in the subsequent motion. [2]
3. Find the time taken from the instant when \(A\) and \(B\) are 1500 m apart to the instant when \(A\) and \(B\) are at the point of closest approach. [2]

A cyclist starting from rest accelerates uniformly at \(1.5 \text{ m s}^{-2}\) for \(4\) s and then travels at constant speed.

1. Sketch a velocity-time graph to represent the first \(10\) seconds of the cyclist's motion. [2]
2. Calculate the distance travelled by the cyclist in the first \(10\) seconds. [2]

A small ball \(B\) moves in the plane of a fixed horizontal axis \(Ox\), which lies on horizontal ground, and a fixed vertically upwards axis \(Oy\). \(B\) is projected from \(O\) with a velocity whose components along \(Ox\) and \(Oy\) are \(U \mathrm{m s}^{-1}\) and \(V \mathrm{m s}^{-1}\), respectively. The units of \(x\) and \(y\) are metres. \(B\) is modelled as a particle moving freely under gravity.

1. Show that the path of \(B\) has equation \(2U^2 y = 2UVx - gx^2\). [3]

During its motion, \(B\) just clears a vertical wall of height \(\frac{1}{3}a\) m at a horizontal distance \(a\) m from \(O\). \(B\) strikes the ground at a horizontal distance \(3a\) m beyond the wall.

1. Determine the angle of projection of \(B\). Give your answer in degrees correct to 3 significant figures. [5]
2. Given that the speed of projection of \(B\) is \(54.6 \mathrm{m s}^{-1}\), determine the value of \(a\). [2]
3. Hence find the maximum height of \(B\) above the ground during its motion. [3]
4. State one refinement of the model, other than including air resistance, that would make it more realistic. [1]

In this question you should take the acceleration due to gravity to be \(10 \text{ms^{-2}\).} \includegraphics{figure_12} A small ball \(P\) is projected from a point \(A\) with speed \(39 \text{ms}^{-1}\) at an angle of elevation \(\theta\), where \(\sin \theta = \frac{5}{13}\) and \(\cos \theta = \frac{12}{13}\). Point \(A\) is \(20 \text{m}\) vertically above a point \(B\) on horizontal ground. The ball first lands at a point \(C\) on the horizontal ground (see diagram). The ball \(P\) is modelled as a particle moving freely under gravity.

1. Find the maximum height of \(P\) above the ground during its motion. [3]

The time taken for \(P\) to travel from \(A\) to \(C\) is \(7\) seconds.

1. Determine the value of \(T\). [3]
2. State one limitation of the model, other than air resistance or the wind, that could affect the answer to part (b). [1]

At the instant that \(P\) is projected, a second small ball \(Q\) is released from rest at \(B\) and moves towards \(C\) along the horizontal ground. At time \(t\) seconds, where \(t \geq 0\), the velocity \(v \text{ms}^{-1}\) of \(Q\) is given by $$v = kt^3 + 6t^2 + \frac{3}{2}t,$$ where \(k\) is a positive constant.

1. Given that \(P\) and \(Q\) collide at \(C\), determine the acceleration of \(Q\) immediately before this collision. [6]

A ball is released from rest from a height \(h\) metres above horizontal ground and falls freely downwards. When the ball reaches the ground, its speed is \(v\) m s\(^{-1}\), where \(v \leq 10\) Show that $$h \leq \frac{50}{g}$$ [3 marks]

A ball, initially at rest, is dropped from a vertical height of \(h\) metres above the Earth's surface. After 4 seconds the ball's height above the Earth's surface is \(0.2h\) metres.

1. Assuming air resistance can be ignored, show that $$h = 10g$$ [3 marks]
2. Assuming air resistance cannot be ignored, explain the effect that this would have on the value of \(h\) in part (a). [1 mark]

In this question use \(g = 9.8\) m s\(^{-2}\) A ball is launched vertically upwards from the Earth's surface with velocity \(u\) m s\(^{-1}\) The ball reaches a maximum height of 15 metres. You may assume that air resistance can be ignored. Find the value of \(u\) [3 marks]

In this question use \(g = 9.81\) m s\(^{-2}\) A particle is projected with an initial speed \(u\), at an angle of 35° above the horizontal. It lands at a point 10 metres vertically below its starting position. The particle takes 1.5 seconds to reach the highest point of its trajectory.

1. Find \(u\). [3 marks]
2. Find the total time that the particle is in flight. [3 marks]

In a school experiment, a particle, of mass \(m\) kilograms, is released from rest at a point \(h\) metres above the ground. At the instant it reaches the ground, the particle has velocity \(v \text{ m s}^{-1}\)

1. Show that $$v = \sqrt{2gh}$$ [2 marks]
2. A student correctly used \(h = 18\) and measured \(v\) as 20 The student's teacher claims that the machine measuring the velocity must have been faulty. Determine if the teacher's claim is correct. Fully justify your answer. [3 marks]

Two particles \(A\) and \(B\) are released from rest from different starting points above a horizontal surface. \(A\) is released from a height of \(h\) metres. \(B\) is released at a time \(t\) seconds after \(A\) from a height of \(kh\) metres, where \(0 < k < 1\) Both particles land on the surface \(5\) seconds after \(A\) was released. Assuming any resistance forces may be ignored, prove that $$t = 5(1 - \sqrt{k})$$ Fully justify your answer. [5 marks]

In this question use \(g = 9.8\) m s\(^{-2}\) An apple tree stands on horizontal ground. An apple hangs, at rest, from a branch of the tree. A second apple also hangs, at rest, from a different branch of the tree. The vertical distance between the two apples is \(d\) centimetres. At the same instant both apples begin to fall freely under gravity. The first apple hits the ground after 0.5 seconds. The second apple hits the ground 0.1 seconds later. Show that \(d\) is approximately 54 [4 marks]

In this question use \(g = 9.8\) m s\(^{-2}\) A toy shoots balls upwards with an initial velocity of 7 m s\(^{-1}\) The advertisement for this toy claims the balls can reach a maximum height of 2.5 metres from the ground.

1. Suppose that the toy shoots the balls vertically upwards.
   1. Verify the claim in the advertisement. [2 marks]
   2. State two modelling assumptions you have made in verifying this claim. [2 marks]
2. In fact the toy shoots the balls anywhere between 0 and 11 degrees from the vertical. The range of maximum heights, \(h\) metres, above the ground which can be reached by the balls may be expressed as $$k \leq h \leq 2.5$$ Find the value of \(k\) [4 marks]

In this question use \(g = 9.81\) m s\(^{-2}\). A ball is projected from the origin. After 2.5 seconds, the ball lands at the point with position vector \((40\mathbf{i} - 10\mathbf{j})\) metres. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. Assume that there are no resistance forces acting on the ball.

1. Find the speed of the ball when it is at a height of 3 metres above its initial position. [6 marks]
2. State the speed of the ball when it is at its maximum height. [1 mark]
3. Explain why the answer you found in part (b) may not be the actual speed of the ball when it is at its maximum height. [1 mark]

A particle P is projected from a fixed point O with initial velocity \(u\mathbf{i} + ku\mathbf{j}\), where \(k\) is a positive constant. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the horizontal and vertically upward directions respectively. P moves with constant gravitational acceleration of magnitude \(g\). At time \(t \geq 0\), particle P has position vector \(\mathbf{r}\) relative to O.

1. Starting from an expression for \(\ddot{\mathbf{r}}\), use integration to derive the formula $$\mathbf{r} = ut\mathbf{i} + \left(kut - \frac{1}{2}gt^2\right)\mathbf{j}.$$ [4]

The position vector \(\mathbf{r}\) of P at time \(t \geq 0\) can be expressed as \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\), where the axes Ox and Oy are horizontally and vertically upwards through O respectively. The axis Ox lies on horizontal ground.

1. Show that the path of P has cartesian equation $$gy^2 - 2ku^2x + 2u^2y = 0.$$ [3]
2. Hence find, in terms of \(g\), \(k\) and \(u\), the maximum height of P above the ground during its motion. [3]

The maximum height P reaches above the ground is equal to the distance OA, where A is the point where P first hits the ground.

1. Determine the value of \(k\). [3]