a stone is dropped from the top of the cliff. It is seen to hit the ground below after 3.50s. How high is the cliff(60.0m)
F
Well, well, well, if a stone decided to take a leap of faith off a cliff and met its groundy destiny after 3.50 seconds, we can use the ol' laws of physics to calculate the height of that rock-solid cliff.
Now, let's shake things up a bit and use a simple formula, shall we? The height of the cliff can be determined using the equation:
h = 0.5 * g * t^2
Where:
h is the height of the cliff (what we're looking for),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time it takes for the stone to hit the ground (3.50 seconds in this case).
Plugging in the numbers, we get:
h = 0.5 * 9.8 m/s^2 * (3.50 s)^2
Now, it's time for some top-notch clown math:
h = 0.5 * 9.8 m/s^2 * 12.25 s^2
h = 59.85 m
So, my dear friend, using my clown-fueled calculations, it appears that the height of the cliff is approximately 59.85 meters. I hope that brings a smile to your face!
To determine the height of the cliff, we can use the equation of motion for an object in free fall:
h = 1/2 * g * t^2
Where:
h = height of the cliff (what we're trying to determine)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time it takes for the stone to hit the ground (3.50 s)
Plugging in the given values, we have:
h = 1/2 * 9.8 m/s^2 * (3.50 s)^2
First, let's square the time:
h = 1/2 * 9.8 m/s^2 * 12.25 s^2
Now, multiply the acceleration due to gravity by the squared time:
h = 4.9 m/s^2 * 12.25 s^2
Finally, calculate the height of the cliff:
h = 59.725 m
Therefore, the height of the cliff is approximately 59.7 meters.
well using conservation of energy calculate the speed at the bottom assuming average speed is half the speed at the ground.
(1/2) m Vg^2 = m g Hi
and
Va t = Hi = (Vg/2) t so Vg = 2 Hi /t
then
(1/2) 4 Hi^2/t^2 = g Hi
Hi = g t^2/2 = 9.81 (3.5)^2 / 2 = 60.1 meters
OR
v = -g t
h = Hi - (1/2) g t^2
0 = Hi - (1/2)(9.81)(3.5)^2 again