A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 23 m/s. The cliff is h = 54 m above the water's surface. How long does it take for the stone to fall to the water?
With what speed does it strike the water?
If the initial velocity is horizontal, it has not influence on the time of fall.
Time of fall is determined by the height, and by gravity.
h=1/2 g t^2
solve for t, time.
Thank you.
How would I calculate the speed?
the vertical speed is
vf= g *time
the horizontal speed is the kicking speed.
Vftotal=sqrt (Vh^2+ Vvf^2)
To solve this problem, we can use the equations of motion for an object in free fall. We'll also assume that air resistance is negligible.
1. First, let's find the time it takes for the stone to fall to the water's surface. We can use the equation:
h = (1/2) * g * t^2
where h is the height of the cliff (54 m) and t is the time it takes for the stone to fall. g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Rearranging the equation to solve for t, we have:
t = sqrt((2h) / g)
Substituting the given values, we get:
t = sqrt((2 * 54) / 9.8)
≈ sqrt(108 / 9.8)
≈ sqrt(11.02)
≈ 3.32 seconds
Therefore, it takes approximately 3.32 seconds for the stone to fall to the water.
2. Next, let's find the speed at which the stone strikes the water. We can use the equation:
v = g * t
where v is the final velocity (speed) and t is the time it takes for the stone to fall.
Substituting the value of t we found earlier, we have:
v = 9.8 * 3.32
≈ 32.576 m/s
Therefore, the stone strikes the water with a speed of approximately 32.576 m/s.