An inquisitive physics student and mountian climber climbs a 43.6 m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.35 s apart and observes that they cause a single splash. The first stone has an initial velocity of 1.8 m/s. How long after release of the first stone do the two stones hit the water

Explanation:

What we are basically looking for here is how long it takes the first stone to hit the water. We have everything we need to figure that out. We will use the equation

Δx = . Filling in, we will solve for t, the time is takes the first stone to hit the water (which is the same for both since they both hit the water at the same time):

which is a quadratic that we will have to factor. Get it into standard form, setting it equal to 0:

and factor to get that

t = 3.2 s and t = -2.8 s

Since time can't ever be negative, it takes 3.2 s for the stones to hit the water.