
Projectile Motion
A ball is thrown with initial velocity 20 m/s at an angle of elevation of arctan (4/3). Suppose that the ball is thrown up a road inclined at arctan(1/5) to the horizontal. Show that the ball si about 9 m above the road when it reaches its greatest height and the time of flight is 2.72 seconds, and find, correct to the nearest tenth of a metre, the distance the ball has been thrown up the road.
I've already found out from previous parts that:
* the parabolic path of the ball has parametric equationx x=12t and y=16t5t^2
* the horizontal range of the ball is 38.4 m
* the greatest height is 12.8 m

Consider an inclined plane of inclination α. Let a projectile be fired making an angle θ along the horizontal. Call axis along the inclined plane to be xaxis. Thus the velocity makes an angle (θ  α) along xaxis.
The time of flight is given by
$\displaystyle T = \frac{2v^2\sin(\theta  \alpha)}{g\cos\alpha}$
Maximum height above the road is
$\displaystyle h_{max} = \frac{v^2\sin^2(\theta  \alpha)}{2g}$