If someone drops a ball from the top of a building at an initial height of 80 meters, what is the height of the ball to the nearest tenth of a meter at exactly 3 seconds after the ball is...

If someone drops a ball from the top of a building at an initial height of 80 meters, what is the height of the ball to the nearest tenth of a meter at exactly 3 seconds after the ball is released.? How many seconds after the ball is released will it hit the ground?

Asked on by taylorruth

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ishpiro | College Teacher | (Level 1) Educator

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To solve this problem, use the equation of motion. If h is the height of the ball after 3 seconds, then 

`h = h_0 +v_0*t - (g*t^2)/2`  

Here, h_0 is the initial height, 80 meters, v_0 is the initial velocity, which is zero because the ball is dropped, not thrown, g is the acceleration of the free fall, 9.8 m/s^2, and t is time, 3 seconds.

Plugging these values in the equations, get 

`h=80 - (9.8*3^2)/2 = 35.9 m`

The height of the ball 3 seconds after it is released is 35.9 meters.

When the ball hits the ground, its height will be h = 0. Then, time until the ball hits the ground can be found from the equation

`0=80-(9.8*t^2)/2`

From here, `t = sqrt((2*80)/9.8) s`

It will take the ball time t = 4.04 seconds to hit the ground.

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