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

Answer:

1.25 s

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=$\frac{x}{2}$

We know that

$s=ut+\frac{1}{2}gt^2$

Where $g=10m/s^2$

Substitute the values

$\frac{x}{2}=0\times 0.883+\frac{1}{2}(10)(0.883)^2$

$\frac{x}{2}=3.898$

$x=3.898\times 2=7.796$m

Again using the above formula

$7.796=0(t)+\frac{1}{2}(10)t^2$

$7.796=5t^2$

$t^2=\frac{7.796}{5}=1.5592$

$t=\sqrt{1.5592}=1.25 s$

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