Josh Beckett is now looking to throw a ball as far as he possibly can. So this time he is going to throw the ball away from himself at an angle of 45 degrees and at a velocity of 35 m/s. What is the highest point the ball reaches, and how far does the ball go if you assume that the ball is released from and comes to rest at a vertical height of 0m.

Question

anonymous · Answer

The velocity up and horizontal are vectors. If they are 45 degree vectors then they are each 35/sqtr(2). So you need to find out how high that will go.

v^2 = u^2 + 2 as. find s

When it stops going up v will be zero. u you already know, a = gravity (-9.81), and so now you can find s = the height.

Horizontally.... you may find the same equation helpful.