h=1/2 g t^2 solve for time t to hit the surface.
speed? ?
in the vertical v= at
in the horizontal v=13m/s
vf= sqrt(vvertical^2+13^2)
A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 13 m/s. The cliff is h = 58 m above the water's surface, as shown below.
(a) How long does it take for the stone to fall to the water?
s
(b) With what speed does it strike the water?
m/s
2 answers
Assume the initial kick is in a horizontal direction. Vertical motion is not affected.
(a)
(1/2) g t^2 = 58 m when it hits the water
Solve for t.
(b) The new kinetic energy will be the initial kinetic energy PLUS M g H
where H = 58 m.
(M/2)*Vfinal^2 = (M/2)*vi^2 + M g H
Cancel the M's and solve for Vfinal
Vfinal^2 = vi^2 + 2 g H
(a)
(1/2) g t^2 = 58 m when it hits the water
Solve for t.
(b) The new kinetic energy will be the initial kinetic energy PLUS M g H
where H = 58 m.
(M/2)*Vfinal^2 = (M/2)*vi^2 + M g H
Cancel the M's and solve for Vfinal
Vfinal^2 = vi^2 + 2 g H
1 answer