A stone was dropped off a cliff and hit the ground with a speed of 112 ft/s. What is the height of the cliff? (Use 32 ft/s2 for the acceleration due to gravity.)
s = 1/2 at^2
v = at, so t = v/a
s = 1/2 a(v^2/a^2) = 1/2 v^2/a = 112^2/64 = 196 ft
To find the height of the cliff, we can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 ft/s, as the stone hits the ground)
u = initial velocity (112 ft/s, as given)
a = acceleration due to gravity (-32 ft/s^2, as the stone is dropped)
s = displacement (height of the cliff, what we need to find)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (0^2 - 112^2) / (2*(-32))
s = (-12544) / (-64)
s = 196 ft
Therefore, the height of the cliff is 196 feet.
To find the height of the cliff, we need to use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement. The equation is:
v^2 = u^2 + 2as
Where:
v = final velocity (112 ft/s)
u = initial velocity (0 ft/s since the stone was dropped)
a = acceleration (acceleration due to gravity = 32 ft/s^2)
s = displacement (height of the cliff)
We can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
Now let's substitute the given values into the equation and solve for s:
s = (112^2 - 0^2) / (2 * 32)
s = (12544 - 0) / 64
s = 12544 / 64
s ≈ 196 ft
Therefore, the height of the cliff is approximately 196 ft.