A ball is dropped from the top of a tall 80m building and after one second another ball is thrown vertically downward from the same building at 15m/s. Will the second ball overtake the first ball? If so, at what height from the top of the building will the two balls be side by side?
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A ball is dropped from the top of the building that is 80 m tall. Use the formula: s = u*t + (1/2)*a*t^2 where s is the distance traveled in time t, a is the rate of acceleration and u is the initial velocity. For the ball that is dropped, the initial velocity is 0.
80 = (1/2)*10*t^2
=> t = 4 s
The time taken by the next ball to reach the ground is given by 80 = 15*t + 5t^2. The positive root of the quadratic equation is (sqrt 73-3)/2 ~~ 2.77
As the second ball takes less than 3 seconds to reach the ground it overtakes the first ball.
Let the two balls be at the same height t seconds after the first is dropped. 5*t^2 = 15*(t-1) + 5(t-1)^2
=> 5*t^2 = 15*t - 15 + 5t^2 + 5 - 10t
=> 0 = 5*t - 10
=> t = 2
The distance of the balls from the top of the building at this moment of time is 20 m.
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