A stone is thrown vertically upward at a speed of 42.80 m/s at time t=0. A second stone is thrown upward with

A stone is thrown vertically upward at a speed of 42.80 m/s at time t=0. A second stone is thrown upward with the same speed 3.730 seconds later. At what time are the two stones at the same height?

At what height do the two stones pass each other?


What is the upward speed of the second stone as they pass each other?

First stone equation:
H1 = - 1/2 g t^2 + V t
H1 = - 1/2 x 9.8 t^2 + 42.8 t
H1 = -4.9 t^2 + 42.8 t

Second stone:
replace t with (t-3.73)
H2 = -4.9 (t-3.73)^2 + 42.8 (t-3.73)
H2 = -4.9 t^2 + 36.554 t - 68.17321 + 42.8 t - 159.644
H2 = -4.9 t^2 + 79.354 t - 227.817

2 stone pass each other means: H1 = H2
-4.9 t^2 + 42.8 t = -4.9 t^2 + 79.354 t - 227.817
36.554t = 227.817
t = 6.2323 sec (after the first stone thrown)

H1 = -4.9 t^2 + 42.8 t = 76.419 m

V1 = V - gt = 42.8 - 9.8 t
V2 = 42.8 - 9.8 (t - 3.73) = 18.28 m/s