A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler's hand?

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u) / t ... Equation 1

Making t the subject of the equation

t = (v-u) g ... Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s (at the maximum height), u = 8.20 m/s, g = - 9.8 m/s²

t = (0-8.20) / -9.8

t = - 8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2 (0.84)

T = 1.68 s.

T = 1.68 s