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.
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.