Question

A cliff diver jumps off a cliff 45.1 meters above the water. If he must land 17.8

meters from the edge of the cliff to be safe, how fast must he be running before
jumping off the cliff?

Answer 1

Speed = 5.87m/s

Step-by-step explanation:

Given the following data;

Height of cliff = 45.1m

Horizontal distance = 17.8m

We know that acceleration due to gravity is 9.8m/²

Initial velocity = 0m/s

To find the time, we would use the second equation of motion;

`S = ut + \frac {1}{2}at^(2)`

Substituting into the equation, we have;

`45.1 = 0(t) + \frac {1}{2}*9.8*t^(2)`

`45.1 = 0 + 4.9*t^(2)`

`45.1 = 4.9*t^(2)`

`t^(2) = \frac {45.1}{4.9}`

`t = √(9.204)`

t = 3.034 secs

To find the speed;

Mathematically, speed is given by the equation;

`Speed = (distance)/(time)`

Substituting into the above equation;

`S = (17.8)/(3.034)`

Speed = 5.87m/s

Therefore, the cliff diver must be running at 5.87 meters per seconds before jumping off the cliff.