A student is standing on top of the school building (which they shouldn’t be doing) and has a height of 78.4m. They kick a rock off the edge of the building with a velocity of 6 m/s.
How long does it take the rock to fall?
How far away from the base of the building does the rock land?
flight time (t)
78.4 = 1/2 * g * t^2 ... t = √(2 * 78.4 / g)
horizontal distance = 6 m/s * t
To find out how long it takes the rock to fall, you can use the equation of motion for freely falling objects:
h = (1/2) * g * t^2
Where:
h is the height of the building (78.4 m)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken to fall
First, we need to rearrange the equation to solve for t:
t = sqrt(2h / g)
Plugging in the values:
t = sqrt(2 * 78.4 / 9.8)
t ≈ sqrt(16) since 2 * 78.4 / 9.8 = 16
t ≈ 4 seconds
Therefore, it takes approximately 4 seconds for the rock to fall.
Now, to find out how far away from the base of the building the rock lands, you can use the equation for horizontal motion:
d = v * t
Where:
d is the distance traveled
v is the horizontal velocity (equal to the initial velocity of the rock, 6 m/s)
t is the time taken to fall (4 seconds)
Plugging in the values:
d = 6 * 4
d = 24 meters
Therefore, the rock lands approximately 24 meters away from the base of the building.