Question

A rock is held steady over a cliff and dropped. 2.4 seconds later, another rock is thrown straight down at a speed of 7.1 m/s, and hits the first rock. How far have the rocks dropped before they collide? How long is the first rock in the air before it gets hit by the second rock?

Answer 1

equation of motion for both objects, call down the positive direction

y1(t) = 1/2 gt^2

y2(t) = 7.1(t - 2.4) + 1/2 g(t-2.4)^2

notice the use of t-2.4 in the equation of motion for the second object; this takes into account the fact that the second stone was dropped 2.4 second later

they hit when y1=y2:

1/2 gt^2 = 7.1(t-2.4) + 1/2 g(t-2.4)^2

4.9t^2 = 7.1t - 17.04 + 4.9t^2 - 23.52t + 28.224

11.184 = 16.42 t

solve for t: t= 0.68s

after 0.68 s, the first rock has traveled = 1/2 x9.8m/s/s x 0.68^2 = 2.27m