
A ball is thrown upwards
Hey I am doing some exam questions from a mechanics textbook and I am really confused I was just wondering if anyone could help me out The question is:
A ball is thrown upwards at 7ms from a point A which is 8m vertically above horizontal ground. Given that the ball moves freely under gravity find:
the greatest height above the ground attained by the ball
the speed of the ball the instant it when it strikes the ground
Thanks alot :)
A ball is thrown vertically upwards and takes 3 seconds to reach its highest point. Find the times at which the ball is 39.2m above its point of projection.

Define the origin to be at ground level and up to be the positive direction. Let the position be called s, velocity v and acceleration a.
We have a = 9.81 m/s/s due to gravity.
$\displaystyle s_0=8, v_0=7$
equation 1. $\displaystyle s=s_0+v_0t+0.5at^2$
equation 2. $\displaystyle v=v_0+at$
Now when the ball is at its highest it is neither going up nor down, that is v=0. So set equation 2 equal to zero and solve it for t, the time at which the height is maximum. Substitute this value of t into equation 1 to find the maximum height.
For the second part set equation 1 equal to zero and solve for t, the time at which the ball hits the ground. Substitute this into equation 2 to find the velocity when the ball hits the ground. Since we defined the positive direction to be up, the answer will be negative.
For your second question $\displaystyle s_0=0$, v_0 is not known. Use equation 2 with t=3 to find v_0, then you can find the required answer using equation 1 by solving for the only unknown (t).
