Asked by happy
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. how long does it take the stone to hit the ground?
b. With what speed and direction does it strike the ground?
Answers
Answered by
Jai
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`
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`
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