Question

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.883 s. How much time passes during his entire trip from the top down to the water

Answer 1

1.25 s

Step-by-step explanation:

We are given that

Initial velocity of boy=u=0

Time=0.883 s

Let x be the total distance covered by boy in entire trip.

s=
`(x)/(2)`

We know that

`s=ut+(1)/(2)gt^2`

Where
`g=10m/s^2`

Substitute the values

`(x)/(2)=0* 0.883+(1)/(2)(10)(0.883)^2`

`(x)/(2)=3.898`

`x=3.898* 2=7.796`m

Again using the above formula

`7.796=0(t)+(1)/(2)(10)t^2`

`7.796=5t^2`

`t^2=(7.796)/(5)=1.5592`

`t=√(1.5592)=1.25 s`

Hence, 1.25 s passes during his entire tripe from the top down to the water.