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.
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.
1 Answer
4.77 m/s towards thrower
Explanation:
-
Break up initial velocity into components.
Initial velocity horizontal = 14.1cos(45)
Initial velocity vertical = 14.1sin(45) -
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 -
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 -
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 -
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.