Question

A juggler throws a bowling pin straight up with an initial speed of 6.00 m/s m/s . How much time elapses until the bowling pin returns to the juggler's hand?

Answer 1

Using 2nd kinematic equation in vertical direction:

h = U*t + (1/2)*a*t^2

h = vertical displacement of bowling pin, when it returns into the juggler's hand = 0 m

U = Initial velocity of bowling pin = +6.00 m/s (+ve sign since upward)

a = acceleration due to gravity = -g = -9.81 m/s^2 (-ve sign since acceleration due to gravity is always downward)

So,

0 = 6.00*t + (1/2)*(-9.81)*t^2

(1/2)*9.81*t^2 = 6.00*t

t = 6.00*2/9.81

t = 1.22 sec

So ball will return into juggler's hand after 1.22 sec

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