Question

A 0.145-kg baseball pitched horizontally at 36.7 m/s strikes a bat and is popped straight up to a height of 52.5 m . Part B Calculate the direction of the average force on the ball during the contact. Express your answer using two significant figures.

Part A If the contact time is 3.00 ms , calculate the magnitude of the average force on the ball during the contact. Express your answer using two significant figures.

Answer 1

Solution) m = 0.145 kg

U = 36.7 m/s ( horizontally )

U = - 36.7(i) m/s

H = 52.5 m

V = (2gH)^(1/2)

V = (2×9.8×52.5)^(1/2)

V = 32.07 m/s

V = 32.07(j) m/s

(A) t = 3 ms = 3×10^(-3) s

Force , F = ?

F = ((m)(V - U))/t

F = ((0.145)(32.07(j) - (-36.7(i)))/(3×10^(-3))

F = 1773.8(i) + 1550(j) N

Magnitude of force , | F | = ((1773.8)^2 + (1550)^2)^(1/2)

| F | = 2355.60 N

| F | = 2.3×10^(3) N

(B) Direction , theeta = tan inverse(1550/1773.8)

Theeta =   41.1°