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 equation of motion in the y-direction:
h = initial height + (initial vertical velocity * time) - (1/2 * acceleration due to gravity * time^2)
Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s. And at the ground, the final height (h) is 0 m. Therefore, the equation becomes:
0 = 30.1 m + (0 m/s * time) - (1/2 * 9.8 m/s^2 * time^2)
Simplifying the equation:
0 = 30.1 m - 4.9 m/s^2 * time^2
Rearranging the equation:
4.9 m/s^2 * time^2 = 30.1 m
Now, divide both sides of the equation by 4.9 m/s^2 to solve for time:
time^2 = 30.1 m / 4.9 m/s^2
time^2 = 6.1224 s^2
Taking the square root of both sides:
time = √6.1224 s^2
time ≈ 2.474 s
Therefore, the time the ball is in motion is approximately 2.474 seconds.