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A juggler throws a bowling pin straight up with an initial speed of 9.20m/s. How much time elapses before the pin reaches the juggler's hands?

Verified Answer :

\(v=v(0)+at\)

\(9.20=0+(9.81*t)\) \(|\)9.81 because the fall is due to gravity.

\(t= \frac{9.20}{9.81}\)

\(t=0.94seconds\)

Answer:

1.87 s

Explanation:

Initial speed of throw = 9.20 m/s

Net vertical displacement = 0

The bowling pin would be in free fall i.e. a = 9.8 m/s²

Use the second equation of motion:

s = ut + 0.5at²

0 = (9.20)t-0.5(9.8)(t²)

9.20 = 4.9 t

⇒t = 1.87 s

Thus, the total time of flight, the time elapses before the bowling pin falls in juggler's hand is 1.87 s.

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