how high does a stone rise if it initially thrown at a rate of 15m/s?
It rises to a height H such that
M g H = (1/2)M Vo^2
(potential energy gain = kinetic energy loss)
Solving for H,
H = Vo^2/(2 g) = 11.5 m
Vo is the intitial speed
M g H = (1/2)M Vo^2
(potential energy gain = kinetic energy loss)
Solving for H,
H = Vo^2/(2 g) = 11.5 m
Vo is the intitial speed