Question

Suppose that you lob the ball with an initial speed of v = 16.1 m/s, at an angle of θ = 51.9° above the horizontal. At this instant your opponent is d = 11.2 m away from the ball. He begins moving away from you 0.400 s later, hoping to reach the ball and hit it back at the moment that it is h = 2.06 m above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)

Answer 1

I think I have this set up, one second...

Answer 2

ok

Answer 3

find the x component of velocity =vcos(51.9 degrees)
find how long the ball is in the air before its hit back---> 2.06=vsin(51.9degrees)t-4.9t^2

solve for t.

using the value of t you found multiply that by the x component of velocity. this is the distance the ball traveled from you.

subtract the other person's distance, 11.2m, from this distance---> this is the distance the guy has to go to reach the ball in time.

take t and add .4 seconds to it. now take the distance the guy has to travel and divide by (t+.4) --> this is the speed the guy has to travel.