A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30â above horizontal with a speed of 52.0 m/s. If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Horizontal component, Fx =  -7045.59 N (Leftwards)

Vertical component, Fy =  2982.96 N (Upwards)

Explanation:

Mass of the baseball = 0.145 kg

Before impact, v₁ = 40 m/s

After impact, v₂ = 52 m/s

time of impact, t = 1.75 ms = 0.00175 s

According to the second law of motion, F = m(v₂ - v₁)/t

Resolving vectorially,

v₁ = 40i m/s ( right = +ve x)

v₂ = -52 cos 30 i + 52 sin 30 j m/s ( to the left (-ve x) and upwards (+ve y))

v₂ = -45.03i + 36j m/s

F = m(v₂ - v₁)/t

F = 0.145 (-45.03i + 36j - 40i)/ 0.00175

F = 0.145 (-85.03i + 36j)/ 0.00175

F = 82.86 (-85.03i + 36j)

F = -7045.59i + 2982.96j N

Fx =  -7045.59 N (Leftwards)

Fy =  2982.96 N (Upwards)

it has to be b, about 97 percent.