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?...

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?

Expert Answers
justaguide eNotes educator| Certified Educator

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.