Question

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.(Answer in m/s)

Answer 1

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$= 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

$V_f^2-V^2=-2gh$

$V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m$

$V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2$

$V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}$

$V_f=42.5m/s$