Speed of the rock just before it reaches the water 23.0 m below the point where the rock left your hand is 49 m/s
Step-by-step explanation:
After 9 seconds displacement is zero
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = ?
Acceleration, a = -9.81 m/s²
Time, t = 9 s
Displacement, s = 0
Substituting
s = ut + 0.5 at²
0 = u x 9 + 0.5 x -9.81 x 9²
u = 44.15 m /s
Initial velocity is 44.15 m/s
Now we need to find speed when displacement is -23 m
We have equation of motion v² = u² + 2as
Initial velocity, u = 44.15 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = -23 m
Substituting
v² = u² + 2as
v² = 44.15² + 2 x -9.81 x -23
v = 49 m/s
Distance traveled before stopping is 54.45 m
Speed of the rock just before it reaches the water 23.0 m below the point where the rock left your hand is 49 m/s