3.02i Projectile motion: constant acceleration model

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]

In this question take \(g = 10\). A small ball P is projected with speed \(20 \text{ m s}^{-1}\) at an angle of elevation of \((\alpha + \theta)\) from a point O at the bottom of a smooth plane inclined at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{5}{12}\) and \(\tan \theta = \frac{3}{4}\). The ball subsequently hits the plane at a point A, where OA is a line of greatest slope of the plane, as shown in the diagram. \includegraphics{figure_9}

1. Determine the following, in either order.
   [9]

After P hits the plane at A it continues to move away from O. Immediately after hitting the plane at A the direction of motion of P makes an angle \(\beta\) with the horizontal.

1. Determine the maximum possible value of \(\beta\), giving your answer to the nearest degree. [3]

A particle P of mass 1 kg is fixed to one end of a light inextensible string of length 0.5 m. The other end of the string is attached to a fixed point O, which is 1.75 m above a horizontal plane. P is held with the string horizontal and taut. P is then projected vertically downwards with a speed of \(3.2 \text{ m s}^{-1}\).

1. Find the tangential acceleration of P when OP makes an angle of \(20°\) with the horizontal. [2]

The string breaks when the tension in it is 32 N. At this point the angle between OP and the horizontal is \(\theta\).

1. Show that \(\theta = 23.1°\), correct to 1 decimal place. [5]

Particle P subsequently hits the plane at a point A.

1. Determine the speed of P when it arrives at A. [4]
2. Show that A is almost vertically below O. [5]

In this question take \(g = 10\). A smooth ball of mass 0.1 kg is projected from a point on smooth horizontal ground with speed 65 m s\(^{-1}\) at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \frac{3}{4}\). While it is in the air the ball is modelled as a particle moving freely under gravity. The ball bounces on the ground repeatedly. The coefficient of restitution for the first bounce is 0.4.

1. Show that the ball leaves the ground after the first bounce with a horizontal speed of 52 m s\(^{-1}\) and a vertical speed of 15.6 m s\(^{-1}\). Explain your reasoning carefully. [4]
2. Calculate the magnitude of the impulse exerted on the ball by the ground at the first bounce. [2]

Each subsequent bounce is modelled by assuming that the coefficient of restitution is 0.4 and that the bounce takes no time. The ball is in the air for \(T_1\) seconds between projection and bouncing the first time, \(T_2\) seconds between the first and second bounces, and \(T_n\) seconds between the \((n-1)\)th and \(n\)th bounces.

1. 1. Show that \(T_1 = \frac{39}{5}\). [2]
   2. Find an expression for \(T_n\) in terms of \(n\). [2]
2. According to the model, how far does the ball travel horizontally while it is still bouncing? [3]
3. According to the model, what is the motion of the ball after it has stopped bouncing? [1]

Fig. 12 shows \(x\)- and \(y\)- coordinate axes with origin O and the trajectory of a particle projected from O with speed 28 m s\(^{-1}\) at an angle \(\alpha\) to the horizontal. After \(t\) seconds, the particle has horizontal and vertical displacements \(x\) m and \(y\) m. Air resistance should be neglected. \includegraphics{figure_12}

1. Show that the equation of the trajectory is given by $$\tan^2\alpha - \frac{160}{x}\tan\alpha + \frac{160y}{x^2} + 1 = 0.$$ (*) [5]
2. 1. Show that if (*) is treated as an equation with \(\tan\alpha\) as a variable and with \(x\) and \(y\) as constants, then (*) has two distinct real roots for \(\tan\alpha\) when \(y < 40 - \frac{x^2}{160}\). [3]
   2. Show the inequality in part (ii)(A) as a locus on the graph of \(y = 40 - \frac{x^2}{160}\) in the Printed Answer Booklet and label it R. [1]

S is the locus of points \((x, y)\) where (*) has one real root for \(\tan\alpha\). T is the locus of points \((x, y)\) where (*) has no real roots for \(\tan\alpha\).

1. Indicate S and T on the graph in the Printed Answer Booklet. [2]
2. State the significance of R, S and T for the possible trajectories of the particle. [3]

A machine can fire a tennis ball from ground level with a maximum speed of 28 m s\(^{-1}\).

1. State, with a reason, whether a tennis ball fired from the machine can achieve a range of 80 m. [1]

Points \(A\) and \(B\) lie on horizontal ground. At time \(t = 0\) seconds, an object \(P\) is projected from \(A\) towards \(B\) such that \(AB\) is the range of \(P\). The speed of projection is \(24 \cdot 5\) ms\(^{-1}\) in a direction which is 30° above the horizontal.

1. Calculate the range \(AB\) of the object \(P\). [5]

At time \(t = 1\) second, another object \(Q\) is projected from \(B\) towards \(A\) with the same speed of projection \(24 \cdot 5\) ms\(^{-1}\) and in a direction which is also 30° above the horizontal.

1. Determine the height above the ground at which \(P\) and \(Q\) collide. [5]

A tennis ball is projected with velocity vector \((30\mathbf{i} - 14\mathbf{j})\) ms\(^{-1}\) from a point \(P\) which is at a height of \(2.4\) m vertically above a horizontal tennis court. The ball then passes over a net of height \(0.9\) m, before hitting the ground after \(\frac{4}{7}\) s. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. The origin \(O\) lies on the ground directly below the point \(P\). The base of the net is \(x\) m from \(O\). \includegraphics{figure_10}

1. Find the speed of the ball when it first hits the ground, giving your answer correct to one decimal place. [3]
2. After \(\frac{2}{5}\) s, the ball is directly above the net.
   1. Find the position vector of the ball after \(\frac{2}{5}\) s.
   2. Hence determine the value of \(x\) and show that the ball clears the net by approximately \(16\) cm. [4]
3. In fact, the ball clears the net by only \(4\) cm.
   1. Explain why the observed value is different from the value calculated in (b)(ii).
   2. Suggest a possible improvement to this model. [2]

\includegraphics{figure_13} A golfer hits a ball from a point \(A\) with a speed of \(25\text{ms}^{-1}\) at an angle of \(15°\) above the horizontal. While the ball is in the air, it is modelled as a particle moving under the influence of gravity. Take the acceleration due to gravity to be \(10\text{ms}^{-2}\). The ball first lands at a point \(B\) which is \(4\text{m}\) below the level of \(A\) (see diagram).

1. Determine the time taken for the ball to travel from \(A\) to \(B\). [3]
2. Determine the horizontal distance of \(B\) from \(A\). [2]
3. Determine the direction of motion of the ball 1.5 seconds after the golfer hits the ball. [4]

In this question you must show detailed reasoning. \includegraphics{figure_11} A football \(P\) is kicked with speed \(25\,\text{m}\,\text{s}^{-1}\) at an angle of elevation \(\alpha\) from a point \(A\) on horizontal ground. At the same instant a second football \(Q\) is kicked with speed \(15\,\text{m}\,\text{s}^{-1}\) at an angle of elevation \(2\alpha\) from a point \(B\) on the same horizontal ground, where \(AB = 72\) m. The footballs are modelled as particles moving freely under gravity in the same vertical plane and they collide with each other at the point \(C\) (see diagram).

1. Calculate the height of \(C\) above the ground. [7]
2. Find the direction of motion of \(P\) at the moment of impact. [4]
3. Suggest one improvement that could be made to the model. [1]

A ball \(B\) is projected with speed \(V\) at an angle \(\alpha\) above the horizontal from a point \(O\) on horizontal ground. The greatest height of \(B\) above \(O\) is \(H\) and the horizontal range of \(B\) is \(R\). The ball is modelled as a particle moving freely under gravity.

1. Show that
   1. \(H = \frac{V^2}{2g}\sin^2 \alpha\), [2]
   2. \(R = \frac{V^2}{g}\sin 2\alpha\). [3]
2. Hence show that \(16H^2 - 8R_0 H + R^2 = 0\), where \(R_0\) is the maximum range for the given speed of projection. [5]
3. Given that \(R_0 = 200\text{m}\) and \(R = 192\text{m}\), find
   1. the two possible values of the greatest height of \(B\), [2]
   2. the corresponding values of the angle of projection. [3]
4. State one limitation of the model that could affect your answers to part (iii). [1]

A girl is practising netball. She throws the ball from a height of 1.5 m above horizontal ground and aims to get the ball through a hoop. The hoop is 2.5 m vertically above the ground and is 6 m horizontally from the point of projection. The situation is modelled as follows.

• The initial velocity of the ball has magnitude \(U\) m s\(^{-1}\).
• The angle of projection is \(40°\).
• The ball is modelled as a particle.
• The hoop is modelled as a point.

This is shown on the diagram below. \includegraphics{figure_12}

1. For \(U = 10\), find
   1. the greatest height above the ground reached by the ball [5]
   2. the distance between the ball and the hoop when the ball is vertically above the hoop. [4]
2. Calculate the value of \(U\) which allows her to hit the hoop. [3]
3. How appropriate is this model for predicting the path of the ball when it is thrown by the girl? [1]
4. Suggest one improvement that might be made to this model. [1]

\includegraphics{figure_11} A projectile is fired from a point \(O\) in a horizontal plane, with initial speed \(V\), at an angle \(\theta\) to the horizontal (see diagram).

1. Show that the range of the projectile on the horizontal plane is $$\frac{2V^2 \sin \theta \cos \theta}{g}.$$ [4]

There are two vertical walls, each of height \(h\), at distances 30 m and 70 m, respectively, from \(O\) with bases on the horizontal plane. The value of \(\theta\) is \(45°\).

1. If the projectile just clears both walls, state the range of the projectile. [1]
2. Hence find the value of \(V\) and of \(h\). [5]

A tennis ball is served horizontally at a speed of 24 m s\(^{-1}\) from a height of 2.45 m above the ground.

1. Show that it will clear the net at a point where the net is 1 m high and 12 m from the server. [5]
2. How far beyond the net will it land? [4]

It is given that the trajectory of a projectile which is launched with speed \(V\) at an angle \(\alpha\) above the horizontal has equation $$y = x\tan\alpha - \frac{gx^2}{2V^2}(1 + \tan^2\alpha),$$ where the point of projection is the origin, and the \(x\)- and \(y\)-axes are horizontal and vertically upwards respectively.

1. Express the above equation as a quadratic equation in \(\tan\alpha\) and deduce that the boundary of all accessible points for this projectile has equation $$y = \frac{1}{2gV^2}(V^4 - g^2x^2).$$ [4]
2. A stone is thrown with speed \(\sqrt{gh}\) from the top of a vertical tower, of height \(h\), which stands on horizontal ground. Find
   1. the maximum distance, from the foot of the tower, at which the stone can land, [3]
   2. the angle of elevation at which the stone must be thrown to achieve this maximum distance. [3]

A particle is projected with speed \(U \text{ m s}^{-1}\) at an angle \(\theta\) above the horizontal, where \(\sin \theta = \frac{12}{13}\), and reaches its maximum height after \(2.4\) seconds.

1. Find \(U\) and the maximum height reached by the particle. [4]
2. Find the horizontal range of the particle. [4]

A particle is projected with velocity \(V\) at an angle \(\alpha\) to the horizontal up a plane inclined at \(\beta\) to the horizontal, where \(\alpha > \beta\).

1. Show that the time of flight is \(\frac{2V \sin(\alpha - \beta)}{g \cos \beta}\). [3]
2. Show that the range on the inclined plane is \(\frac{2V^2 \sin(\alpha - \beta) \cos \alpha}{g \cos^2 \beta}\). [4]
3. If the particle strikes the plane at right angles, prove that \(\tan \alpha = \cot \beta + 2 \tan \beta\). [5]

A gunner fires one shell from each of two guns on a stationary ship towards a vertical cliff \(AB\) of height \(100\) m whose foot \(A\) is at a horizontal distance \(600\) m from the point of projection.

1. Given that the shell from the first gun hits the cliff, travelling horizontally, at a point \(45\) m above \(A\), determine the initial velocity of the shell. Express your answer in the form \(a\mathbf{i} + b\mathbf{j}\), where \(a\) and \(b\) are integers. [6]
2. The shell from the second gun hits the cliff at its top point \(B\). Given that the initial speed of the shell is \(300\) m s\(^{-1}\), determine the possible angles of projection. [5]

4 A particle \(P\) is projected with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(40 ^ { \circ }\) above the horizontal from a point \(O\) on horizontal ground.

1. Find the height of \(P\) above the ground when \(P\) has speed \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
2. Calculate the length of time for which the speed of \(P\) is less than \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and find the horizontal distance travelled by \(P\) during this time.

3 A ball is projected horizontally with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the top of a tower which is 30 m high. The tower stands on horizontal ground.

1. Find the speed and direction of motion of the ball when it reaches the ground.
2. Calculate the distance from the foot of the tower to the point where the ball reaches the ground.