A stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. Using the fact that acceleration due to gravity is −32ft/s2, what is the height (in feet) of the cliff?
a = -32
v = -32t + c
since it was 'dropped' , when t=0 , v =0
so c = 0
v = -32t
120 = -32t
t= 120/-32 = -15/4
s = -16t^2 + k
when it hit the ground, t = 15/4 , s = 0
0 = -16(225/16) = k
k = 225
s = -16t^2 + 225
So the height of the cliff is 225 ft
Minor note, although your final answer is correct, I would not that the time is not going to be negative because that final velocity is negative just like the acceleration, hence why the problem mentions the "speed" of the object. This ensures you have a positive time, because that is the only thing that makes sense in this situation logically. Good work though!
Ah, the tragic tale of the falling stone! Let's see if we can calculate the height of that cliff with a little humor thrown in, shall we?
To find the height of the cliff, we can use the kinematic equation:
v² = u² + 2as
Where:
- v is the final velocity (in this case, 0 ft/s since the stone hits the ground).
- u is the initial velocity (in this case, 120 ft/s since that's the speed at which the stone was dropped).
- a is the acceleration due to gravity (-32 ft/s², you know, because gravity just pulls everything down).
- s is the displacement (the height of the cliff, what we're trying to find).
Okay, enough chit-chat. Plugging in the values, we can rearrange the equation to solve for s:
0² = (120 ft/s)² + 2(-32 ft/s²)s
Now, be gentle with me, dear human, as I crank the mathy gears in my clown brain.
0 = 14400 ft²/s² - 64 ft/s² s
Rearranging a bit more...
64 ft/s² s = 14400 ft²/s²
Dividing both sides by 64 ft/s²...
s = 225 ft
Aha! Drumroll, please! *insert drumroll sound effect here*
The height of the cliff is 225 feet! Give that stone an extra round of applause for the epic fall!
To find the height of the cliff, we need to use the equation of motion:
v^2 = u^2 + 2as
where
v = final velocity (0 ft/s because the stone hits the ground and comes to rest)
u = initial velocity (120 ft/s)
a = acceleration due to gravity (-32 ft/s^2)
s = displacement (height of the cliff)
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the given values, the calculation becomes:
s = (0^2 - 120^2) / (2 * -32)
s = (-14400) / (-64)
s = 225
Therefore, the height of the cliff is 225 feet.
To calculate the height of the cliff, we can use the kinematic equation for displacement:
s = ut + (1/2)at^2
Where:
s = displacement (height of the cliff)
u = initial velocity (0 ft/s because the stone was dropped)
t = time taken to hit the ground (unknown)
a = acceleration due to gravity (-32 ft/s^2)
Rearranging the equation, we can solve for t:
s = (1/2)at^2
2s/a = t^2
√(2s/a) = t
Substituting the given values:
√(2s/-32) = t
Since the stone was dropped, its initial velocity (u) is 0, and the time taken to hit the ground is the same as the time taken to fall, so we can set u = 0.
Then, the equation becomes:
t = √(-2s/a)
Now we can substitute the values given in the problem:
t = √(-2s/-32)
t = √(s/16)
The stone's speed, 120 ft/s, is equal to the final velocity when it hits the ground. We can find the final velocity (v) using the equation:
v = u + at
Since the stone was dropped, the initial velocity (u) is 0, and we know the acceleration (a) is -32 ft/s^2, so the equation becomes:
v = -32t
Substituting the value of t from the earlier equation:
120 ft/s = -32(√(s/16))
120 = -32√(s/16)
-120/32 = √(s/16)
-3.75 = √(s/16)
Now we can square both sides to eliminate the square root:
(-3.75)^2 = (s/16)
14.06 = s/16
s = 14.06 × 16
s = 224.96 ft
Therefore, the height of the cliff is approximately 224.96 feet.