Question

How fast, and in which direction, must the person run to catch the ball at the level from which it was thrown?

A ball is thrown at 14.1 m/s at 45° above the horizontal. Someone located 30m away along line of the path starts to run just as the ball is thrown.

The answer is 4.77 m/s toward thrower. But i don't know how to get started with it.

Answer 1

4.77 m/s towards thrower

Explanation:

1. Break up initial velocity into components.
   Initial velocity horizontal = 14.1cos(45)
   Initial velocity vertical = 14.1sin(45)

2. Find time the ball is in the air
   At the top of the ball's pathway, it has 0 vertical velocity. therefore final vertical velocity is 0.
   9.8 = `(v-14.1sin45)/t`
   Then multiply the time by two so that you find the total time the ball is in the air.
   t= 2.0347 s

3. Find how far away the ball is thrown
   Using horizontal velocity, find displacement
   `s=vt`
   `s= 14.1cos(45)*2.0347`
   s= 20.29 m

4. The catcher is 30 m away, so he needs to run forwards (or towards the ball) to catch it
   so find how far he must run forwards 30-20.29 = 9.71 m

5. find his velocity using the fact that he runs as soon as the ball is hit
   `v=s/t`
   `v=9.71/2.0347`
   v= 4.77 m/s towards the thrower.