Question

A mountain climber stands at the top of a 47.0-m cliff that overhangs a calm pool of water. She throws two stones vertically downward 1.00 s apart and observes that they cause a single splash. The first stone had an initial velocity of −1.40 m/s. (Indicate the direction with the sign of your answers.) (a) How long after release of the first stone did the two stones hit the water? (Round your answer to at least two decimal places.)

Answer 1

Answer:

t = 2.96 s

Explanation:

Since the two stones hit the water at same instant of time

so we will have

$d =vt + \frac{1}{2}gt^2$

here we know that

d = 47 m

v = 1.4 m/s

$g = 9.81 m/s^2$

$d = 1.40 t + \frac{1}{2}(9.81) t^2$

now by solving above equation for  d = 47 m

t = 2.96 s