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A ball is thrown horizontally off of a building 67 m above the ground. When the ball hits the ground, it

has traveled a horizontal distance of 19 m.
Determine the speed at which the ball was moving after being thrown. Ignore air resistance and use 9.8
m/s² for acceleration due to gravity.

1 Answer

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Final Answer:

The speed at which the ball was moving after being thrown horizontally off the building can be determined using the formula v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height from which the ball is thrown. Substituting the given values, the speed is approximately 34 m/s.

Step-by-step explanation:

When a ball is thrown horizontally, its initial vertical velocity is zero. The only force acting on it is the force of gravity, leading to vertical motion. The vertical distance traveled by the ball can be determined using the equation h = (1/2)gt², where h is the height, g is the acceleration due to gravity, and t is the time of flight. In this case, the height is 67 m.

The time of flight (t) can be found using the horizontal motion equation d = vt, where d is the horizontal distance traveled. Given d = 19 m, and v is the initial horizontal velocity, which is the same as the final horizontal velocity, as there is no horizontal acceleration. Solving for t, we get t = d/v.

Substituting t into the equation for vertical motion and rearranging to solve for v, we get v = √(2gh). Substituting the given values (g = 9.8 m/s² and h = 67 m), we find v to be approximately 34 m/s.

In summary, the speed of the ball after being thrown horizontally off the building is calculated using the equation v = √(2gh), considering the vertical motion and the acceleration due to gravity. The calculated speed is approximately 34 m/s.

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