A ball is thrown horizontally from the roof of a building 33.0 m tall and lands 20.5 m from the base. What was the ball's initial speed?
To find the initial speed of the ball, we can use the horizontal distance traveled and the height of the building.
Let's assume that the initial speed of the ball was "v" and the time it takes for the ball to reach the ground is "t".
Since the ball is thrown horizontally, there is no vertical acceleration acting on it. This means that the time it takes to reach the ground can be found using the formula:
h = (1/2) * g * t^2
Where:
h = height of the building = 33.0 m
g = acceleration due to gravity = 9.8 m/s^2
Substituting the given values, we get:
33.0 = (1/2) * 9.8 * t^2
66.0 = 9.8 * t^2
t^2 = 66.0 / 9.8
t^2 ≈ 6.735 n
Taking the square root of both sides, we find:
t ≈ √(6.735) ≈ 2.6 s
Now that we have determined the time it takes for the ball to reach the ground, we can use it to calculate the horizontal speed. Since the distance traveled is given as 20.5 m, we can use the formula:
v = d / t
Where:
v = initial speed of the ball
d = horizontal distance traveled = 20.5 m
t = time taken to reach the ground ≈ 2.6 s
Substituting the given values, we get:
v = 20.5 / 2.6
v ≈ 7.88 m/s
Therefore, the ball's initial speed was approximately 7.88 m/s.