Two possible trajectories through point

5 A particle is projected from a point \(O\) on horizontal ground. The velocity of projection has magnitude \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and direction upwards at an angle \(\theta\) to the horizontal. The particle passes through the point which is 7 m above the ground and 16 m horizontally from \(O\), and hits the ground at the point \(A\).

1. Using the equation of the particle's trajectory and the identity \(\sec ^ { 2 } \theta = 1 + \tan ^ { 2 } \theta\), show that the possible values of \(\tan \theta\) are \(\frac { 3 } { 4 }\) and \(\frac { 17 } { 4 }\).
2. Find the distance \(O A\) for each of the two possible values of \(\tan \theta\).
3. Sketch in the same diagram the two possible trajectories.

3 Two particles \(P\) and \(Q\) are projected simultaneously with speed \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) on a horizontal plane. Both particles subsequently pass at different times through the point \(A\) which has horizontal and vertically upward displacements from \(O\) of 40 m and 15 m respectively.

1. By considering the equation of the trajectory of a projectile, show that each angle of projection satisfies the equation \(\tan ^ { 2 } \theta - 8 \tan \theta + 4 = 0\).
2. Calculate the distance between the points at which \(P\) and \(Q\) strike the plane.

7 A particle \(P\) is projected with speed \(35 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) on a horizontal plane. In the subsequent motion, the horizontal and vertically upwards displacements of \(P\) from \(O\) are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively. The equation of the trajectory of \(P\) is $$y = k x - \frac { \left( 1 + k ^ { 2 } \right) x ^ { 2 } } { 245 }$$ where \(k\) is a constant. \(P\) passes through the points \(A ( 14 , a )\) and \(B ( 42,2 a )\), where \(a\) is a constant.

1. Calculate the two possible values of \(k\) and hence show that the larger of the two possible angles of projection is \(63.435 ^ { \circ }\), correct to 3 decimal places. For the larger angle of projection, calculate
2. the time after projection when \(P\) passes through \(A\),
3. the speed and direction of motion of \(P\) when it passes through \(B\). {www.cie.org.uk} after the live examination series. }

8 A particle is projected with speed \(49 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) 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 \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Express \(x\) and \(y\) in terms of \(\theta\) and \(t\), and hence show that $$y = x \tan \theta - \frac { x ^ { 2 } \left( 1 + \tan ^ { 2 } \theta \right) } { 490 } .$$
   \includegraphics[max width=\textwidth, alt={}]{35477eb8-59e0-4de6-889c-1f5841f65eec-4_627_1249_1699_447}
   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).
2. Find \(\theta _ { 1 }\) and \(\theta _ { 2 }\).
3. Calculate the distance between \(A _ { 1 }\) and \(A _ { 2 }\).

5. \begin{figure}[h] \end{figure} A small ball is projected with speed \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) on horizontal ground. After moving for \(T\) seconds, the ball passes through the point \(A\). The point \(A\) is 40 m horizontally and 20 m vertically from the point \(O\), as shown in Figure 2. The motion of the ball from \(O\) to \(A\) is modelled as that of a particle moving freely under gravity. Given that the ball is projected at an angle \(\alpha\) to the ground, use the model to

1. show that \(T = \frac { 10 } { 7 \cos \alpha }\)
2. show that \(\tan ^ { 2 } \alpha - 4 \tan \alpha + 3 = 0\)
3. find the greatest possible height, in metres, of the ball above the ground as the ball moves from \(O\) to \(A\). The model does not include air resistance.
4. State one other limitation of the model.

6 A particle is projected with speed \(v \mathrm {~ms} ^ { - 1 }\) from a point \(O\) on horizontal ground. The angle of projection is \(\theta ^ { \circ }\) above the horizontal. At time \(t\) seconds after the instant of projection the horizontal displacement of the particle from \(O\) is \(x \mathrm {~m}\) and the upward vertical displacement from \(O\) is \(y \mathrm {~m}\).

1. Show that $$y = x \tan \theta - \frac { 4.9 x ^ { 2 } } { v ^ { 2 } \cos ^ { 2 } \theta } .$$ A stone is thrown from the top of a vertical cliff 100 m high. The initial speed of the stone is \(16 \mathrm {~ms} ^ { - 1 }\) and the angle of projection is \(\theta ^ { \circ }\) to the horizontal. The stone hits the sea 40 m from the foot of the cliff.
2. Find the two possible values of \(\theta\). \includegraphics[max width=\textwidth, alt={}, center]{8492ec9b-3327-4d89-aaa4-bf98cdf0ebdc-3_623_995_1475_536} A uniform ladder \(A B\) of weight \(W \mathrm {~N}\) and length 4 m rests with its end \(A\) on rough horizontal ground and its end \(B\) against a smooth vertical wall. The ladder is inclined at an angle \(\theta\) to the horizontal where \(\tan \theta = \frac { 1 } { 2 }\) (see diagram). A small object \(S\) of weight \(2 W \mathrm {~N}\) is placed on the ladder at a point \(C\), which is 1 m from \(A\). The coefficient of friction between the ladder and the ground is \(\mu\) and the system is in limiting equilibrium.

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

1. Write down the equation of the trajectory, in terms of \(\tan \theta\).
2. The particle passes through a point whose horizontal and vertical distances from \(O\) are 72 m and \(y \mathrm {~m}\) respectively. By considering the equation of the trajectory as a quadratic equation in \(\tan \theta\), or otherwise, find the greatest possible value of \(y\).

A particle is projected from a point \(O\) on horizontal ground. The velocity of projection has magnitude \(20 \text{ m s}^{-1}\) and direction upwards at an angle \(\theta\) to the horizontal. The particle passes through the point which is 7 m above the ground and 16 m horizontally from \(O\), and hits the ground at the point \(A\).

1. Using the equation of the particle's trajectory and the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), show that the possible values of \(\tan \theta\) are \(\frac{4}{3}\) and \(\frac{1}{4}\). [4]
2. Find the distance \(OA\) for each of the two possible values of \(\tan \theta\). [3]
3. Sketch in the same diagram the two possible trajectories. [2]

A particle \(P\) is projected with speed \(u\) at an angle \(\alpha\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.

1. Derive the equation of the trajectory of \(P\) in the form $$y = x\tan\alpha - \frac{gx^2}{2u^2}\sec^2\alpha.$$ [3]

The point \(Q\) is the highest point on the trajectory of \(P\) in the case where \(\alpha = 45°\).

1. Show that the \(x\)-coordinate of \(Q\) is \(\frac{u^2}{2g}\). [3]
2. Find the other value of \(\alpha\) for which \(P\) would pass through the point \(Q\). [4]

A particle \(P\) is projected with speed \(u\) m s\(^{-1}\) at an angle of \(\theta\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) s are denoted by \(x\) m and \(y\) m respectively.

1. Show that the equation of the trajectory is given by $$y = x \tan \theta - \frac{gx^2}{2u^2}(1 + \tan^2 \theta).$$ [4]

In the subsequent motion \(P\) passes through the point with coordinates \((30, 20)\).

1. Given that one possible value of \(\tan \theta\) is \(\frac{4}{3}\), find the other possible value of \(\tan \theta\). [5]

\includegraphics{figure_4} A small ball is projected from a fixed point \(O\) so as to hit a target \(T\) which is at a horizontal distance \(9a\) from \(O\) and at a height \(6a\) above the level of \(O\). The ball is projected with speed \(\sqrt{(27ag)}\) at an angle \(\theta\) to the horizontal, as shown in Figure 4. The ball is modelled as a particle moving freely under gravity.

1. Show that tan\(^2 \theta - 6\) tan \(\theta + 5 = 0\) [7]

The two possible angles of projection are \(\theta_1\) and \(\theta_2\), where \(\theta_1 > \theta_2\).

1. Find tan \(\theta_1\) and tan \(\theta_2\). [3]

The particle is projected at the larger angle \(\theta_1\).

1. Show that the time of flight from \(O\) to \(T\) is \(\sqrt{\left(\frac{78a}{g}\right)}\). [3]
2. Find the speed of the particle immediately before it hits \(T\). [3]

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]

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]