A stone is thrown vertically upward at a speed of 27.70 m/s at time t=0. A second stone is thrown upward with the same speed 2.140 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 downward speed of the first stone as they pass each other?
H1 = 27.7 t - 4.9 t^2
H2 = 27.7 (t-2) - 4.9 (t-2)^2
so
27.7t-4.9t^2=27.7t-55.4-4.9(t^2-4t+4)
0 = -55.4 + 19.6 t - 19.6
19.6 t = 75
t = 3.83 seconds
h = 27.7(3.83) - 4.9(3.83)2
h = 34.21
v = Vi - g t
v = 27.7 - 9.81 (3.83)
v = -9.87
so speed down = 9.87