An inquisitive physics student and mountain climber climbs a 54.0-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 1.82 m/s.
(a) how long after release of the first stone do the two stones hit the water? s
(b) what initial velocity must the second stone have if the two stones are to hit the water simultaneously? magnitude m/s direction
(c) what is the speed of each stone at the instant the two stones hit the water? first stone m/s second stone m/s

a) After the release of the first stone the two stones hit the water at 3.14 seconds.

b)Initial velocity of second stone = 14.74 m/s

c) Speed of first stone at the instant the two stones hit the water = 32.62 m/s

Speed of second stone at the instant the two stones hit the water = 35.73 m/s

Explanation:

a) For first stone we have

  Initial velocity, u =  1.82 m/s

  Acceleration , a = 9.81 m/s²

  Displacement, s = 54 m

  We have equation of motion s= ut + 0.5 at²

 Substituting

   s= ut + 0.5 at²

    54 = 1.82 x t + 0.5 x 9.81 x t²

    4.905 t² + 1.82 t - 54 = 0

    t = 3.14 s or t = -3.51 s  (not possible)

After the release of the first stone the two stones hit the water at 3.14 seconds.

b) The second stone is thrown after 1 s

 Time taken by second stone to reach top of pool of water = 3.14 - 1 = 2.14 s

  Time, t =  2.14 m/s

  Acceleration , a = 9.81 m/s²

  Displacement, s = 54 m

  We have equation of motion s= ut + 0.5 at²

 Substituting

   s= ut + 0.5 at²

    54 = u x 2.14 + 0.5 x 9.81 x 2.14²

    u = 14.74 m/s

  Initial velocity of second stone = 14.74 m/s

c) We have equation of motion v = u + at

 For first stone

                  Initial velocity, u =  1.82 m/s

                  Acceleration , a = 9.81 m/s²

                  Time, t = 3.14 s

                  v = 1.82 + 9.81 x 3.14 = 32.62 m/s

 For second stone

                  Initial velocity, u =  14.74 m/s

                  Acceleration , a = 9.81 m/s²

                  Time, t = 2.14 s

                  v = 14.74 + 9.81 x 2.14 = 35.73 m/s

Speed of first stone at the instant the two stones hit the water = 32.62 m/s

Speed of second stone at the instant the two stones hit the water = 35.73 m/s

answer: what are you talking about?

explanation:

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.