Question

When Reggie stepped up to the plate and hit a 0.150-kg fast ball traveling
at 36.0 m/s, the impact caused the ball to leave his bat with a velocity of
45.0 m/s in the opposite direction.
a. If the impact lasted for 0.002 s, what force did Reggie exert on the baseball?
b. Suppose the ball had instead travelled upwards at an angle of 30°. Find the force Reggie
exerted on the baseball in vector form. Then find the magnitude of the force Reggie
exerted.
I need help with B

Answer 1

a. V = Vo + a*t.
-45 = 36 + a*0.002, a = ?

F = M*a.

b. V = Vo + a*t.
45[30o] = 36 + a*0.002,
39 + 22.5i = 36 + a*0.002,
3 + 22.5i = a*0.002,
a = 1500 + 11,250i = 11,350 m/s^2[82.4o].

F = M*a = 0.15 * 11,350[82.4o] = 1,702N[82.4o].

Magnitude = 1,702 N.

Answer 2

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