Question

As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 19.0 m/s. In the process, he moves his hand through a distance of 1.10 m. If the ball has a mass of 0.150 kg, find the force he exerts on the ball to give it this upward speed.

Answer 1

Answer:

$F_{player}=26.1N$

Explanation:

Given data

Initial speed vi=0 m/s

Final speed vf=19.0 m/s

Δy=1.10 m

Mass of ball m=0.150 kg

To find

Force

Solution

First we need to find acceleration of the ball.So we can apply:

vf²=vi²+2aΔy

vf²=0+2aΔy

$a=\frac{(19.0m/s)^2}{2*1.10m}\\ a=164.1m/s^2$

Because we get the acceleration of the ball.We can now find the force the players exerts on the ball by using Newtons second law

∑F=F.player -mg=ma

$F_{player}-mg=ma\\F_{player}=mg+ma\\F_{player}=m[g+a]\\F_{player}=0.150kg[9.81+164.1]\\F_{player}=26.1N$