A stone is dropped from the edge of a roof, and hits the ground with a velocity of -145 feet per second. How high (in feet) is the roof
Ke = (1/2) m v^2 = (1/2) m (145^2)
pe = m g h where g = 32.2 in ft/seconds units
32.2 m h = (1/2) m (145^2)
h = 326 feet
vf^2=2gh
solve for h, in feet
g= 32ft/s^2
Hey Bob, I was wondering why gravity is 32m/s^2 instead of 9.8m/s^2. Can you explain why this is?
Check the units m/s^2 and ft/s^2
To find the height of the roof, we need to determine the distance the stone has fallen before hitting the ground.
We can use the equation of motion for free fall:
h = (1/2)gt^2
Where:
h is the height in feet
g is the acceleration due to gravity, approximately 32.2 ft/s^2
t is the time in seconds
In this case, we need to find the time it takes for the stone to hit the ground. We can use the equation for final velocity of an object in free fall:
v = u + gt
Where:
v is the final velocity (-145 ft/s)
u is the initial velocity (0 ft/s, as the stone is dropped)
g is the acceleration due to gravity (32.2 ft/s^2)
t is the time in seconds
Rearranging the equation, we have:
t = (v - u) / g
Plugging in the values, we get:
t = (-145 ft/s - 0 ft/s) / 32.2 ft/s^2
t = -145 ft/s / 32.2 ft/s^2
t ≈ -4.5 seconds
Since time cannot be negative in this context, we take the absolute value:
t ≈ 4.5 seconds
Now, we can substitute the value of t into the equation for height:
h = (1/2)gt^2
h = (1/2)(32.2 ft/s^2)(4.5 s)^2
h = (1/2)(32.2 ft/s^2)(20.25 s^2)
h ≈ 326.96 feet
Therefore, the height of the roof is approximately 326.96 feet.