A person standing on top of a 40-m high cliff throws a rock with a mass of 0.25 kg off the cliff.The rock leaves the person's hand with a speed of 11 m/s and makes an angle of 75 degrees above the horizontal.What is the kinetic of the rock when it hits the ground at the bottom of the cliff?

Question

Quizplus · Accepted Answer

The potential energy of the rock at the top of the cliff is given by: 
PE = mgh = (0.25 kg)(9.81 m/s^2)(40 m) = 98.1 J

The velocity of the rock can be broken down into its horizontal and vertical components:
vx = v*cos(75) = 2.65 m/s
vy = v*sin(75) = 10.72 m/s

The time it takes for the rock to hit the ground can be found using:
t = (2*vy) / g = (2*10.72 m/s) / 9.81 m/s^2 = 2.19 s

The horizontal distance the rock travels before hitting the ground can be found using:
x = vx*t = (2.65 m/s)*(2.19 s) = 5.81 m

The kinetic energy of the rock just before hitting the ground is given by:
KE = (1/2)mv^2 = (1/2)*(0.25 kg)*(11 m/s)^2 = 15.125 J

Therefore, the total energy of the rock just before hitting the ground is given by:
Total Energy = KE + PE = 98.1 J + 15.125 J = 113.225 J

Rounding to the nearest whole number, the kinetic energy of the rock just before hitting the ground is 113 J.

A Person Standing on Top of a 40-M High Cliff