Suppose you roll a bowling ball at 2.5 m/s across the roof of a flat building. It leaves the edge and strikes the ground 2.0 s later.
How high is the roof?
How far from the edge of the roof does the ball hit the ground?
To calculate the height of the roof, we can use the formula for the vertical motion of an object:
h = v0*t + (1/2)*a*t^2
Where:
h = height of the roof
v0 = initial vertical velocity
t = time taken for the ball to hit the ground
a = acceleration due to gravity (-9.8 m/s^2)
Given:
v0 = 0 (since the ball is released from rest)
t = 2.0 s
a = -9.8 m/s^2
Substituting the values into the formula:
h = 0 + (1/2)*(-9.8)*(2.0)^2
h = 0 - 9.8*4.0
h = -39.2 meters
Since the height cannot be negative (assuming the ground level is taken as the reference point), we can conclude that the roof is 39.2 meters high.
Now, to calculate the horizontal distance from the edge of the roof to the point where the ball hits the ground, we can use the formula for horizontal motion:
d = v*t
Where:
d = horizontal distance
v = horizontal velocity
t = time taken for the ball to hit the ground
Given:
v = 2.5 m/s
t = 2.0 s
Substituting the values into the formula:
d = 2.5 * 2.0
d = 5 meters
Therefore, the ball hit the ground 5 meters away from the edge of the roof.