Question

A mountain climber stands at the top of a 33.5-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

-2.40 m/s. (Indicate the direction with the sign of your
answer.)(a) How long after release of the first stone did the two
stones hit the water?

s

(b) What initial velocity must the second stone have had, given
that they hit the water simultaneously?

m/s

(c) What was the velocity of each stone at the instant it hit the
water?

first stone | m/s |

second stone | m/s |

Answer #1

Let velocity of second stone be v.

Time taken for first stone to hit water = t.

THen we have, -33.5 = -2.4t+0.5*-g*t^2, where g=9.8

Solving, we have t=2.4s

So, after 2.4 seconds of release of the first stone,the two stones hit the water

Stone 2 hits water 1 second faster, so time for second stone = 1.4 s

Equation of motion: -33.5 = u*1.4 + 0.5*-9.8*1.4^2

we have u=-17.06 m/s

So initial velocity of stone 2 is -17.06 m/s

v=u+at

For first stone, v=-2.4-9.8*2.4=-25.92 m/s

For second stone, v=-17.06-9.8*1.4=-30.78 m/s

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