Two stones are thrown from the top edge of a building with a speed of 20 m/s, one straight down and the other straight up. The first one hits the street in a time t1. How much later is it before the second stone hits

Second stone hist at 4.08 s later.

Explanation:

The second stone straight up, so we need to find the rise and descent time.

At the top, the final speed is 0 and t is the rise time.

Let's recall that the rise time is equal to the descent time.

So the total time of the stone will be

Now, the second stone hits the street in t2 = 4.08 s + t1 s

But we just need the time after the first stone hist the street. So we just subtract the time t1 to the second time t2, which means t2 = 4.08 s.

Therefore stone hist at 4.08 s later.

more than, less

less than, more

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