An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff t

Question

An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 1.84 m/s.

(a) How long after release of the first stone do the two stones hit the water?

(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?

(c) What is the speed of each stone at the instant the two stones hit the water?

Explanation / Answer

(a) As, v2-u2 = 2gh

=> v2 = 2*9.8*51 + 1.84*1.84

           =1002.98

=> v = 31.66 m/sec

As, v=u +gt

=> 31.66 = 1.84 +9.8t

=> t =3.04 sec

after release of the first stone the two stones hit the water after = 3.04 sec

(b) Second stone is thrown after 1 sec

=> time taken by second stone to hit water = 2.04 sec

=> 51 = u*2.04 + 1/2*9.8*2.043*2.043

=> u = 14.97 m/sec

initial velocity must the second stone have if the two stones are to hit the water simultaneously =14.97 m/sec

(c) speed of first stone when it hit the water = 31.66 m/sec

Speed of second stone when it hit the water = 14.97 + 9.8*2.043

                                                                        = 34.99 m/sec

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