8 A ball is kicked from ground level over horizontal ground. It leaves the ground at a speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and at an angle \(\theta\) to the horizontal such that \(\cos \theta = 0.96\) and \(\sin \theta = 0.28\).
- Show that the height, \(y \mathrm {~m}\), of the ball above the ground \(t\) seconds after projection is given by \(y = 7 t - 4.9 t ^ { 2 }\). Show also that the horizontal distance, \(x \mathrm {~m}\), travelled by this time is given by \(x = 24 t\).
- Calculate the maximum height reached by the ball.
- Calculate the times at which the ball is at half its maximum height.
Find the horizontal distance travelled by the ball between these times.
- Determine the following when \(t = 1.25\).
(A) The vertical component of the velocity of the ball.
(B) Whether the ball is rising or falling. (You should give a reason for your answer.)
(C) The speed of the ball. - Show that the equation of the trajectory of the ball is
$$y = \frac { 0.7 x } { 576 } ( 240 - 7 x )$$
Hence, or otherwise, find the range of the ball.