Question

In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 14.0 m/s at an angle of 40.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 48.0 m/s at 30.0° above the horizontal. (Take positive î to be the horizontal direction from the plate toward center field and take positive ĵ to be the upward vertical direction.) (a) Determine the impulse delivered to the ball. (b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases linearly to zero in another 4.00 ms, what is the maximum force on the ball?

Answer 1

let
m = 0.2 kg
v1 = 14 m/s
v2 = 48 m/s

v1x = -14*cos(40) = -10.7 m/s
v1y = -14*sin(40) = -9.0 m/s

v2x = 48*cos(30) = 41.6 m/s
v2y = 48*sin(30) = 24 m/s

a) Impulse in x-direction, Ix = m*(v2x - v1x)

= 0.2*(41.6 - (-10.7))

= 10.46 N.s

Impulse in y-direction, Iy = m*(v2y - v1y)

= 0.2*(24 - (-9.0))

= 6.60 N.s

Impulse delivered to the ball, I = sqrt(Ix^2 + Iy^2)

= sqrt(10.46^2 + 6.6^2)

= 12.4 N.s (or) kg.m/s <<<<<<-------Answer

b) let F_max is the maximum force on the ball.

we know, Impulse = area under F-t curve

I = (1/2)*F_max*delta_t1 + F_max*delta_t2 + (1/2)*F_max*delta_t3

I = F_max*(delta_t1/2 + delta_t2 + delta_t3/2)

F_max = I/(delta_t1/2 + delta_t2 + delta_t3/2)

= 12.4/(4*10^-3/2 + 20*10^-3 + 4*10^-3/2)

= 517 N <<<<<<-------Answer