The coach throws a baseball to a player with a initial speed of 20 m/s at an angle of 45º with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed must the player run to catch the ball at the same height at which it was released?

Question

anonymous · Answer

The answer is 3.19 m/s

But I don't know how to get there... Can someone show me how to get this answer? Thanks.

anonymous · Answer

find the time and distance traveled when its height is zero (back to the same height it started at).  then find the velocity the player must have to cover the distance from his x position (x=50) to the x location where the ball 'lands' in the time it takes the ball to travel to that point.

anonymous · Answer

solve for t:
$$height:  0 = 20*\sin  45 *t - \frac{ g t ^{2} }{ 2 }$$
using t, solve for x
$$distance: x = 20*\cos  45 *t $$

distance - 50 is how far he has to run, and the time you found is how long he has to cover that distance

anonymous · Answer

make sense?

anonymous · Answer

Why is height = 0? We don't know height.

anonymous · Answer

"to catch the ball at the same height at which it was released"

delta h =0

anonymous · Answer

Oh ok. Thanks.

anonymous · Answer

get the right answer now?