A man threw a ball horizontally out of an elevator that was traveling upward at a rate of 5.50m/s. If the ball was thrown at a velocity of 10.5m/s when the elevator was 15.3m above the floor of the atrium. How far from the base of the elevator did it land?
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To find the horizontal distance the ball travels from the base of the elevator, we need to determine the time it takes for the ball to reach the ground.
First, let's calculate the time it takes for the ball to reach the ground based on its vertical motion. We can use the equation:
h = ut + (1/2)gt^2
Where:
h = vertical displacement from the top of the elevator to the ground (-15.3m since the elevator is moving upward)
u = initial vertical velocity (0m/s since the ball is thrown horizontally)
g = acceleration due to gravity (-9.8m/s^2)
Rearranging and substituting the values, we have:
15.3 = 0t + (1/2)(-9.8)t^2
Multiplying through by 2, we get:
30.6 = -4.9t^2
Dividing both sides by -4.9, we have:
t^2 = -30.6/-4.9
t^2 = 6.244
Taking the square root of both sides, we get:
t ≈ √6.244
t ≈ 2.5 seconds (rounded to one decimal place)
Now that we know the time it takes for the ball to reach the ground, we can calculate the horizontal distance it travels. Since the ball was thrown horizontally, it maintains a constant horizontal velocity of 10.5m/s.
The horizontal distance (d) can be determined using the formula:
d = v * t
Where:
v = horizontal velocity (10.5m/s)
t = time (2.5 seconds)
Substituting the values, we have:
d = 10.5 * 2.5
d = 26.25 meters
Therefore, the ball lands approximately 26.25 meters from the base of the elevator.