Projectile Motion

Nov 2009
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=16t-5t^2
* the horizontal range of the ball is 38.4 m
* the greatest height is 12.8 m
Jun 2009
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 x-axis. Thus the velocity makes an angle (θ - α) along x-axis.

The time of flight is given by

\(\displaystyle T = \frac{2v^2\sin(\theta - \alpha)}{g\cos\alpha}\)

Maximum height above the road is

\(\displaystyle h_{max} = \frac{v^2\sin^2(\theta - \alpha)}{2g}\)
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