A 150 g baseball is thrown from the outfield to the third baseman. The throwing

Question

A 150 g baseball is thrown from the outfield to the third baseman. The throwing motion of the arm covers a 2 m distance. The ball accelerates from his hand (from rest) at 150 m/s2, and leaves his hand at an angle (with horizontal) of 30degree . The ball is caught at the same height as thrown. Neglect air resistance and rotational motion of the ball. (Equations of motion, Newton's Laws, energy and momentum)

a)The speed of the thrown baseball is _____ m/s.

b) The ball rises to a height of _____ m.

c) The ball is in the air for _____ s.

d) The total mechanical energy of the baseball in flight is _____ J.

e) The momentum change of the baseball in throwing motion is _____ kgm/s

Explanation / Answer

Here ,

m = 0.150 Kg

distance , d = 2 m

a = 150 m/s^2

theta = 30 degree

a) let the speed is u

Using third equation of motion

u^2 = 2 *a * d

u^2 = 2 * 2 * 150

u = 24.5 m/s

the speed of thrown ball is 24.5 m/s

b)

height of ball = (u * sin(theta))^2/(2 * g)

height of ball = (24.5 * sin(30))^2/(2 * 9.8)

height of ball = 7.66 m

c)time of flight = 2 * u * sin(theta)/g
time of flight = 2 * 24.5 * sin(30)/9.8

time of flight = 2.5 s

the ball is in the air for 2.5 s

d)
total mechanical energy of the baseball = 0.5 * m * u^2

total mechanical energy of the baseball = 0.5 * 0.150 * 24.5^2

total mechanical energy of the baseball = 45.02 J

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