a 2.50 N rock is thrown down from a cliff at 10.0 m/s.

Question

a. what is the velocity after falling 25.0 m?
b. velocity as it hits the water 125 m below?
c. kinetic energy as it hits the H20?

Answer 1

I am not going to do all of these for you. This and the previous one are basically the same.

The mass does not matter until we get to part c

define velocity and distance positive down
Vi = 10
g = 9.81
distance at 0 = 0
v = Vi + g t
d = Vi t + (1/2) g t^2
at 25 meters
25 = 10 t + 4.9 t^2
4.9 t^2 + 10 t -25 = 0
t = [ -10 +/- sqrt(100+ 490)]/9.81
t = [-10 +/- 24.3 ]/9.81
t = 1.46 seconds
so
v = 10 + g t = 24.3 m/s at 25 m

at water
d = 125
125 = 10 t + 4.9 t^2
4.9 t^2 + 10 t - 125 = 0
t = [ -10 +/- sqrt (100+2450)]/9.81
t = 4.13 seconds
so
v = 10 + 9.81(4.13) = 50.5 m/s at 125 m

Ke = (1/2) m v^2
= (1/2)(2.5/9.81)(50.5)^2
= 325 Joules