Question

A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before striking the water

Answer 1

14.05m/s

Step-by-step explanation:

We are given that

Horizontal speed of diver=
`v_x=1.2m/s`

Distance, y=-10 m

It is taken as negative because the diver goes downward.

There is no air resistance therefore, there is acceleration due to gravity .

We know that

`g=-9.8 m/s^2`

Initial vertical velocity,
`u_y=0`

`v^2-u^2=2as`

Using the formula

`v^2_y-0=2* (-9.8)(-10)`

`v_y=√(2(-9.8)(-10))`

`v_y=14 m/s`

`v=√(v^2_x+v^2_y)`

Using the formula

`v=√((1.2)^2+(14)^2)=14.05m/s`

Hence, the speed of diver before just striking the water=14.05m/s