Question

The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 12.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 177 ms. The ball starts from rest.
(a) Through what distance does it move before its release?
(b) What are the magnitude and direction of the force the pitcher exerts on the ball?

I got part a) 1.062 but i cant get part b correct please help

Answer 1

a = change in velocity/time
a = (12-0)/ (177* 10^-3) = 67.8 m/s^2

distance = average velocity * time
= 6 * 177*10^-3 = 1.06 meters

F = m a
m = 2.28/9.81 = .232 kg
F = .232 * 67.8 = 15.7 horizontal force
there is also a vertical force up by the pitcher = 2.28 to keep it horizontal
so
F^2 = 2.28^2 + 15.7^2
angle up T
tan T = 2.28/15.7

Answer 2

sin^-1(y/x)