A diver running 1.0 m/s dives out horizontally from the edge of a vertical cliff and 2.8 s later reaches the water below. How high was the cliff?

Question

Don't understand how to do this problem.

Answer 1

Velocity is a vector quantity, which can be decomposed into vertical and horizontal components.  The horizontal and vertical velocity are independent from eachother, so the diver's horizontal velocity has no effect on his vertical velocity.  This means that the diver's horizontal velocity at the time of his jump has no effect on how fast he falls.

 

So the problem contains useless information.  The only useful piece of information is how long it takes him to reach the water (2.8 s).

 

Since he dives out horizontally, his entire velocity is directed horizontally; that is, the initial vertical velocity is 0.  From the standpoint of his vertical velocity, he is falling from rest.

 

We don't need to consider horizontal motion to solve the problem, so it reduces to motion in 1 dimension (in the vertical direction). 

 

We can use a kinematic equation:

 

s(t) = s₀ + v₀*t + 1/2 * at²

 

We are trying to find s₀, the initial vertical height.  a = -9.8 m/s², the acceleration due to gravity, and v_0, the initial vertical velocity, is 0.  So the equation reduces to

 

s(t) = s₀ - 4.9t².

 

We also know that s(2.8) = 0; that is, taking the water level as a height of zero, the diver's position at 2.8 seconds is at the water level (=0).

 

So,

 

0 = s₀ - 4.9(2.8)², which with a bit of algebra yields

s₀ = 4.9(2.8)² = 38.416 meters.

 

That's a pretty tall cliff!