Question

A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool of water. The climber throws two stones vertically 1.0 s apart and observes that they cause a single splash when they hit the water. The first stone has an initial velocity of +2.0 m/s.
a. How long after release of the first stone will the two stones hit the water?
b. What is the initial velocity of the second stone when it is thrown?
c. What will the velocity of each stone be at the instant both stones hit the water?

Answer 1

Answer:

a. $t=2.997\ s$ is the time after which the two stone hit the water surface.

b. $u_2=1.998\ m.s^{-1}$

c. $v_1=31.3688\ m.s^{-1}$   &  $v_2=31.3686\ m.s^{-1}$

Explanation:

Given:

• height of the cliff, $h=50\ m$

• time gap between the projection of stones, $\Delta t=1\ s$

• initial velocity of first stone, $u_1=+2\ m.s^{-1}$

Here positive sign means that the stone is thrown vertically downward in the direction of gravity.

a.

Using equation of motion:

$h=u_1.t+\frac{1}{2} g.t^2$

$50=2t+4.9t^2$

$t=2.997\ s$ is the time after which the two stone hit the water surface.

b.

using the equation of motion for second stone:

$h=u_2.t+\frac{1}{2} g.t^2$

$50=u_2\times 2.997+4.9\times 2.997^2$

$u_2=1.998\ m.s^{-1}$

c.

Final velocity of the stone at the instant of touching the water surface:

$v_1^2=u_1^2+2.g.h$

$v_1^2=2^2+2\times 9.8\times50$

$v_1=31.3688\ m.s^{-1}$

&

$v_2^2=u_2^2+2g.h$

$v_2^2=1.998^2+2\times 9.8\times 50$

$v_2=31.3686\ m.s^{-1}$