A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts for 1.85 ms, which results in a velocity of 37.0 m/s in the opposite direction. What was the magnitude of the horizontal force

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

F = 0.140(37+ 28.9) /1.85x10^-3

F = 9.226/1.85x10^-3

F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

the answer is b i think