Question

A baseball is thrown from second base to first base in a double play. If the ball is released with a vertical velocity of 3.2 m/s and is caught 18 m away at the same height as it had at the instant of release

Answer #1

Initial vertical velocity of ball = vy = 3.2 m/s

Horizontal distance covered by ball = x = 18 m = Horizontal velocity x time of flight

time of flight = T

Using y = voT + 0.5 aT^2

0 = 3.2T + 0.5(-9.81)T^2

T = 0.65 s

Using

18 m = Horizontal velocity x time of flight

18 = vx ( 0.65)

vx = 27.69 s

Net velocity with which ball was thrown = (vy^2 + vx^2)^0.5 = ( 3.2^2 + 27.69^2)^0.5 = 27.87 m/s

Direction of velocity = arctan (Vy/Vx) = arctan (3.2/27.69) =
6.59^{o} above horizontal

**Ball was thrown with 27.87 m/s velocity at an angle of
6.96 ^{o} with horizontal**

Maximum height reached by ball = y

y = vo(T/2) + 0.5 a(T/2)^2

y = 3.2(0.325) + 0.5 (-9.81)(0.325)^2 [ T/2 = 0.65/2 = 0.325]

y = **0.52 m [ maximum height reached by
ball]**

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