A stone is dropped from the edge of a roof, and hits the ground with a velocity of −170 feet per second. Assume the acceleration due to gravity is -32 feet per second squared. How high (in feet) is the roof?
-170 = -32t
now, s = 16t^2
To find the height of the roof, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity = -170 ft/s
u = initial velocity = 0 ft/s (as the stone is dropped)
a = acceleration = -32 ft/s^2 (negative because it is against the direction of motion)
s = height of the roof (what we want to find)
Plugging in the values into the equation, we have:
(-170)^2 = (0)^2 + 2*(-32)*s
28900 = 0 - 64s
64s = -28900
Dividing both sides of the equation by 64, we get:
s = -28900 / 64
s ≈ -451.56 ft
Since distance cannot be negative, we take the absolute value of the result to get the height of the roof:
s ≈ 451.56 ft
Therefore, the height of the roof is approximately 451.56 feet.
To find the height of the roof, we can use the kinematic equation for displacement:
d = ut + (1/2)at^2
Where:
d = displacement (height of the roof)
u = initial velocity (0 ft/s since the stone was dropped)
t = time taken for the stone to hit the ground
a = acceleration due to gravity (-32 ft/s^2)
First, let's find the time taken for the stone to hit the ground. We can use the equation:
v = u + at
Where:
v = final velocity (-170 ft/s)
u = initial velocity (0 ft/s)
a = acceleration due to gravity (-32 ft/s^2)
Plugging in the values:
-170 ft/s = 0 ft/s + (-32 ft/s^2)t
Rearranging the equation to solve for t:
t = (-170 ft/s) / (-32 ft/s^2)
t ≈ 5.31 seconds
Now we can calculate the height of the roof using the first equation:
d = (0 ft/s)(5.31 s) + (1/2)(-32 ft/s^2)(5.31 s)^2
d ≈ 0 ft + (1/2)(-32 ft/s^2)(28.1761 s^2)
d ≈ -16 ft/s^2 * 28.1761 s^2
d ≈ -450.8186 ft^2/s^2
Since the height cannot be negative, we take the absolute value:
d ≈ 450.8186 ft
Therefore, the height of the roof is approximately 450.82 feet.