Question

A 0.155-kg baseball pitched horizontally at 34.0 m/s strikes a bat and is popped straight up to a height of 35.5 m .Find the direction (degrees) of the average force on the ball.

Answer 1

given

m = 0.155 kg
v1 = 34 m/s
h = 35.5 m/s

speed of ball after hitting with bat, v2 = sqrt(2*g*h)

= sqrt(2*9.8*35.5)

= 26.4 m/s

let us assume initially the ball is moving towards -x direction.
Impulse in x-direction, Jx = m*(v2x - v1x)

= 0.155*(0 - (-34))

= 5.27 N.s

impulse in y-direction, Jy = m*(v2y - v1y)

= 0.155*(26.4 - 0 )

= 4.09

the direction of impulse, theta = tan^-1(Jy/Jx)

= tan^-1(4.09/5.27)

= 37.8 degrees above horizontal

we know, Impulse = F_avg*delta_t

Impulse and F_avg always have the same direction.

so, the direction of force is also 37.8 degrees above horizontal <<<<<<<<=-----------------Answer