Answer:
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
