Asked by Karie

If the juggler wants the balls to remain in the air for 1.4 seconds, (a) what should the vertical velocity be? (b) What should the horizontal velocity be?

The distance between the juggler’s hands is 0.70 meters and the acceleration due to gravity is −9.80 m/s^2.

Answers

Answered by Elena
Δt =1.4 s.
Vertical motion
v(y) = v(yo) -gt
t =0.7 s, v(y) = 0
v(yo) =gt = 9.8•0.7 =6.86 m/s.
Horizontal velocity
v(x) =Δx/Δt =0.7/1.4 =0.5 m/s.
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions