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?
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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
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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