A stone is dropped from a cliff 100 ft. high. Disregarding friction;
a. how long does it take the stone to hit the ground?
b. With what speed and direction does it strike the ground?
a.
The stone is experiencing free fall, and therefore the motion involved is uniformly accelerated motion (UAM). We can use the formula,
h = vo*t - (1/2)gt^2
where
vo = initial velocity
t = time
g = acceleration due to gravity = 32.2 ft/s^2
Since it's free fall, vo = 0. Substituting,
100 = -(0.5)(-32.2)t^2
t = ?
Now solve for t.
b.
We can also use this formula for UAM:
vf^2 - vo^2 = 2gh
where
vf = final velocity
Terminal velocity is what we're looking for, and in this case it's vf since it's the velocity just before it hits the ground. Substituting the values,
vf^2 - 0 = 2(32.2)(100)
Now solve for vf. Note that the direction is downwards.
Hope this helps~ `u`
To find the time it takes for the stone to hit the ground, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = distance (100 ft)
u = initial velocity (0 ft/s since the stone is dropped)
a = acceleration due to gravity (-32.2 ft/s^2, considering downward direction)
t = time
Since the stone is dropped, the initial velocity is zero. So we can simplify the equation as:
s = (1/2)at^2
Plugging in the values:
100 = (1/2)(-32.2)t^2
Rearranging the equation:
t^2 = -[(2 * 100) / (-32.2)]
t^2 = 6.211
t = √6.211
t ≈ 2.49 seconds
Therefore, it takes approximately 2.49 seconds for the stone to hit the ground.
To find the speed and direction with which the stone strikes the ground, we can use another kinematic equation:
v = u + at
where:
v = final velocity
u = initial velocity (0 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2, considering downward direction)
Plugging in the values:
v = 0 + (-32.2)(2.49)
v ≈ -79.97 ft/s (rounded to two decimal places)
The negative sign indicates that the velocity is directed downward. The magnitude of the velocity is approximately 79.97 ft/s, meaning the stone strikes the ground at around 79.97 feet per second downward.