Question

A stone is thrown vertically upward at a speed of 41.50 m/s at time t=0. A second stone is thrown upward with the same speed 3.020 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?

Answer 1

Answer:

t1 = t2 + 3.02         V = 41.5

V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2

Both stones reach the same height after the specified times

V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)

2 V / g = t1 + t2 = 2t1 + 3.02

t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s

t2 = t1 + 3.02 = 5.74 sec

Check:

41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m

41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m

Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s