Hoo! This one sounds complicated. And they didn’t even tell us how high the cliff is! (Doesn’t it matter?) We draw a picture of the problem, as in Fig. It turns out that if we understand something about the motion of pellet A the problem is much simpler. Let’s ask:
What is the velocity v of pellet A when it returns to the height at which it was thrown? Here we don’t care about the time, just the distances and velocities are involved, so we want to use When the pellet returns to the original height then y = 0 and so we get:
and the proper solution to this equation is
Here we choose the minus sign because the pellet is moving downward at that time. So when the pellet returns to the same height it has the same speed but is moving in the opposite direction.
But recall that pellet B was thrown downward with speed 30 m s, that is, its initial velocity was −30 m/s . So from this point on, the motion of pellet A is the same as that of pellet B.
So from that point on it will be the same amount of time until A hits the ground. Therefore the amount of time which A spends in the air above that spent by B is the time it spends it takes to go up and then down to the original height. Therefore we now want to answer the question: How long does it take A to go up and back to the original height?
To answer this question we can use with x = 0. We can also ask how long it take until the velocity equals −30 m/s , and that will be simpler. So using with a = −9.80 m/s2 we solve for t:
We get:
Summing up, it takes 6.1 s for pellet A to go up and back down to the original height; this is the amount of time it spends in the air longer than the time B is in the air. So pellet A hits the ground 6.1 s after B hits the ground.
