a ball is throwm at an angle of 45° to the horizontal wirh kinetic energy k what is the kinetic energy at the highest point of trajectory
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At initial position velocity = v
K = 0.5mv^2
At highest point velocity = vCos45 = v/√2
K’ = 0.5m(v/√2)^2
K’ = 0.5mv^2 / 2
K’ = K/2
Kinetic energy at highest point is K/2
K = 0.5mv^2
At highest point velocity = vCos45 = v/√2
K’ = 0.5m(v/√2)^2
K’ = 0.5mv^2 / 2
K’ = K/2
Kinetic energy at highest point is K/2
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