If a rock is dropped from a height of 64 feet, how long does it take to hit the ground? How fast is it going when it hits the ground?
h=g*t solve for time t
vf=1/2 g t^2 solve for vf
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To answer these questions, we can use equations of motion and the value of acceleration due to gravity, which is approximately 32.2 feet per second squared.
Step 1: Find the time it takes to hit the ground.
The equation to calculate the time taken for an object to fall from a height without any initial velocity is given by:
time = sqrt((2 * height) / acceleration)
Given:
- Height (h) = 64 feet
- Acceleration due to gravity (g) = 32.2 feet/second^2
Substituting these values into the equation, we get:
time = sqrt((2 * 64) / 32.2)
time = sqrt(128 / 32.2)
time = sqrt(3.97515)
time ≈ 1.99 seconds (rounded to two decimal places)
Therefore, it takes approximately 1.99 seconds for the rock to hit the ground.
Step 2: Find the speed of the rock when it hits the ground.
The final velocity of an object falling from a height is given by the equation:
final velocity = initial velocity + (acceleration * time)
Since the rock is being dropped (meaning its initial velocity is 0), the equation simplifies to:
final velocity = acceleration * time
Substituting the known values:
final velocity = 32.2 * 1.99
final velocity ≈ 64.06 feet/second (rounded to two decimal places)
Therefore, the rock is moving at approximately 64.06 feet per second when it hits the ground.
To calculate the time it takes for the rock to hit the ground, we can use the equation of motion for an object in free fall:
h = (1/2) * g * t^2
Where:
h = height of the object (in this case, 64 feet)
g = acceleration due to gravity (approximately 32.2 feet per second squared)
t = time
Now, we can rearrange the equation to solve for time:
t = sqrt(2h / g)
Substituting the given values:
t = sqrt(2 * 64 / 32.2)
t = sqrt(4)
t = 2 seconds
Therefore, it takes 2 seconds for the rock to hit the ground.
Now, to calculate the final velocity of the rock when it hits the ground, we can use the equation:
v = g * t
Where:
v = final velocity
g = acceleration due to gravity (as mentioned before)
t = time (as calculated before)
Substituting the given values:
v = 32.2 * 2
v = 64.4 feet per second
Therefore, the rock is traveling at a speed of 64.4 feet per second when it hits the ground.