Problem 43.

A swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the ledge at the bottom which is 2 m wide and 9 m below the top of the cliff?

Solution:

This is the projectile motion with horizontal initial velocity. In this case the motion in horizontal direction is described by the following equation:

Where v is the initial velocity.

The motion along vertical axis y is described by the following equations (y-component of initial velocity is 0):

We know that the final point of the motion is the end of the ledge. The coordinates of the final point are: x=2 m and y = -9m.

We can substitute these numbers in the above equations and obtain:

From the seconds equation we can find time:

Then we substitute this time in the first equation and obtain the initial velocity: