A ball is thrown horizontally from the top of a building 30.1 m high. The ball strikes the ground at a point 66.8 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s
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To find the time the ball is in motion, we can use the kinematic equation for horizontal motion:
d = v * t
where:
d = horizontal distance (66.8 m)
v = horizontal velocity (which is constant)
t = time
Since the ball is thrown horizontally from the top of the building, the initial vertical velocity is 0 and there is no vertical acceleration. Therefore, the time it takes for the ball to reach the ground is the same as the time it takes to cover the horizontal distance.
To find the horizontal velocity of the ball, we can use the kinematic equation for vertical motion:
y = v0 * t + (1/2) * a * t^2
where:
y = vertical displacement (30.1 m)
v0 = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation, we get:
y = (1/2) * a * t^2
Plugging in the values:
30.1 = (1/2) * (-9.8) * t^2
Simplifying:
t^2 = (2*30.1) / 9.8
t^2 = 61.2 / 9.8
t^2 = 6.2449
Square rooting both sides, we get:
t = √(6.2449)
t ≈ 2.5 s
Therefore, the time the ball is in motion is approximately 2.5 seconds.