A juggler throws a bowling pin straight up with an initial speed of 7.70m/s. How much time elapses until the bowling pin returns to the juggler's hand?

1.57 s

Explanation:

From Newton's equation of motion.

v = u + gt ... Equation 1

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

Given: v = 0 m/s (The pin is being thrown up), u = 7.7 m/s, g = - 9.81 m/s² (against air resistant)

Substituting into equation 1

0 = - 7.7 + (-9.81t)

9.81t = 7.7

t = 7.7/9.81

t = 0.785 s.

Note:Time taken for the pin to return to the juggler's hand is doubled the time taken to reach the maximum height

T = 2t

T = 2 (0.785)

T = 1.57 s.

Hence the time taken for the pin to returns to the juggler's hand = 1.57 s