Since there is no initial vertical velocity component, the time to fall (T) is given by
H = (1/2)gT^2
H is the cliff height.
T = sqrt(2H/g)
The speed Vf at impact can be easily computed with conservation of energy.
(1/2)m Vf^2 =(1/2)m Vo^2 = mgH
The mass cancels out. Vo is the initial velocity (when kicked)
Vf^2 = Vo^2 + 2 g H
A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 20 m/s. The cliff is h = 53 m above the water's surface, as shown below.
(a) How long does it take for the stone to fall to the water?
(b) With what speed does it strike the water?
2 answers
Oh i get it. You have to use the equation for an accelerating straight-line object. Thanks