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?

Answer:1.87 sExplanation:Initial speed of throw = 9.20 m/sNet vertical displacement = 0 The bowling pin would be in free fall i.e. a = 9.8 m/s²Use the second equation of motion:s = ut + 0.5at²0 = (9.20)t-0.5(9.8)(t²)9.20 = 4.9 t ⇒t = 1.87 sThus, the total time of flight, the time elapses before the bowling pin falls in juggler's hand is 1.87 s.

[tex]v=v(0)+at[/tex][tex]9.20=0+(9.81*t)[/tex] [tex]| [/tex]9.81 because the fall is due to gravity.[tex]t= \frac{9.20}{9.81} [/tex][tex]t=0.94seconds[/tex]