Question

A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to

Answer 1

Answer:

20 m

Step-by-step explanation:

We are given that

Initial horizontal speed ,$u_x=15 m/s$

Time, t=2 s

Initial vertical velocity, $u_y=0$

We know that

$h=u_yt+\frac{1}{2}gt^2$

Where $g=-9.8m/s^2$

Using the formula

$h=0(2)+\frac{1}{2}(-9.8)(2)^2$

$h=-19.6 m\approx -20 m$

The negative sign indicates that the displacement of stone  is in downward direction.

Hence, the height of the cliff is closest to 20 m