Question

A stone is thrown vertically upwards with a speed of 24m/s,two seconds later a second stone is dropped from the same point.Find where the two stones meet

Answer 1

A ball is dropped from a height of 5m onto a concrete floor and rebounds with a speed of 0.8 times the downward speed on arrival.Find the height reached from the rebound

Answer 2

Let t be the time after the first stone is thrown, Y1 be the height of the first stone and Y2 the height of the dropped stone. Require that Y1 = Y2 and solve for t.

Y1 = 24t - 4.9 t^2
Y2 = -4.9(t-2)^2 = -4.9 t^2 +19.6t -19.6 (provided t > 2 s)

24t = 19.6t -19.6

There is no solution with t > 2 s
The stones never meet while in free flight