Question

A diver running 2.6 m/s dives out horizontally from the edge of a vertical cliff and 2.6 s later reaches the water below.

How high was the cliff?
How far from its base did the diver hit the water?

Answer 1

Step-by-step explanation:

A diver running 2.6 m/s dives out horizontally from the edge of a vertical cliff and 2.6 s later reaches the water below.

(a) It means in horizontal direction,

Initial velocity is 0 and acceleration is g. Let h is the height of the cliff. Using equation of motion as :

`h=ut+(1)/(2)gt^2\\\\h=(gt^2)/(2)\\\\h=(9.8* 2.6^2)/(2)\\\\h=33.12\ m`

(b) In horizontal direction,

Initial velocity of the driver is 2.6 m/s and acceleration is equal to 0. Let x is the distance from its base did the diver hit the water.

`d=ut+(1)/(2)gt^2\\\\d=ut\\\\d=2.6* 2.6\\\\d=6.76\ m`