Question

A baseball of mass 0.25 kg is hit at home plate with a speed of 40 m/s. When it lands in a seat in the left-field bleachers a horizontal distance 120 m from home plate, it is moving at 30 m/s. If the ball lands 20 m above the spot where it was hit, how much work is done on it by air resistance?

Answer 1

Answer:

the work done by air resistance is 38.5 J

Explanation:

given information:

mass of the ball, m = 0.25 kg

initial speed, $v_{1}$ = 40 m/s

final speed, $v_{2}$ = 30 m/s

horizontal distance, x = 120 m

Δh = 20 m

according to conventional energy

W = $E_{2}$ - $E_{1}$

where

$E_{1}$ is initial energy

$E_{2}$  is final energy

E = KE + PE

KE is kinetic energy

PE is potential energy

W = $E_{2}$ - $E_{1}$

   = mg$h_{2}$ + $\frac{1}{2} mv_{2} ^{2}$ - (mg$h_{1}$ + $\frac{1}{2} mv_{1} ^{2}$)

   = mg($h_{2} -h_{1}$ )+ $\frac{1}{2} m(v_{2} ^{2} - v_{1} ^{2})$

   = m (gΔh + $\frac{1}{2} (v_{2} ^{2} - v_{1} ^{2})$)

   = 0.25 ( (9.8) (20) + $\frac{1}{2} (30 ^{2} - 40 ^{2})$)

   = - 38.5 J