Now assume that the pitcher in part d throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. the batter then hits the ball so it goes directly back to the pitcher along the same straight line. what is the ball's x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4n⋅s to the baseball?

25.9 m/s

Explanation:

mass of ball, m = 0.145 kg

initial velocity, u = + 32 m/s

It bounce back with the velocity but in opposite direction so final velocity,

v = - v

Impulse, I = - 8.4 Ns

Impulse is defined as the change in momentum

I = m v - mu = m (v - u)

- 8.4 = 0.145 x (- v - 3 2)

- 57.9 = - v - 32

v = 57.9 - 32 = 25.9 m/s

Thus, the final speed of the ball is 25.9 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s