Asked by Ash

You have an spring gun. The end of the gun is at 1 meter above the ground. This gun is shoot horizontally with no air resistance. The cannon ball hits the ground 1.32 meters ahead from it was shoot. What is the cannon ball's initial velocity?

Answers

Answered by Henry
h = 0.5g*t^2 = 1 m.
4.9t^2 = 1
t^2 = 0.204
t = o.452 s. = Fall time.

d = Vo*t = 1.32 m.
Vo*0.452 = 1.32
Vo = 2.92 m/s.
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