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Robert, a stuntman, projects himself from a height of 50. 0 meters in a horizontal direction. He safely lands at a horizontal distance of 100. 0 meters. Calculate the horizontal velocity with which robert projected himself. A. 4. 90 meters/second b. 31. 30 meters/second c. 10. 20 meters/second d. 12. 20 meters/second.

1 Answer

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Final answer:

To find the horizontal velocity with which Robert projected himself, the time of flight is calculated using the height of the fall and acceleration due to gravity. Then, using the time of flight and the horizontal distance covered, the horizontal velocity is approximately 31.30 meters/second. So the correct option is b.

Step-by-step explanation:

To calculate the horizontal velocity with which Robert projected himself, we first need to use the given vertical distance (height) to find the flight time. Since the only force acting on him vertically during his free fall is gravity, we can use the equation of motion for constant acceleration (in this case, the acceleration due to gravity which is approximately 9.81 meters/second2). The formula h = 1/2gt2 applies here, where h is the height and g is the acceleration due to gravity.

The given height is 50.0 meters, so the equation becomes:
50.0 = 0.5 * 9.81 * t2,
which results in
t2 = 50.0 / (0.5 * 9.81),
t2 ≈ 10.20,
t ≈ 3.19 seconds (time of flight).

Now, we use the horizontal distance and the time of flight to find the horizontal velocity (v = d/t, where d is distance and t is time). The horizontal distance is given as 100.0 meters.
So, the horizontal velocity is:
v = 100.0 / 3.19 ≈ 31.35 meters/second.

Therefore, the correct answer, rounded to two decimal places, is b. 31.30 meters/second.

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