A 15 kg rock is horizontally from very high cliff at a speed of 65m/s. what is the speed of the rock after it has fallen a vertical distance of 35m?
V^2 = Vo^2 + 2g*h
Vo = 0
2g = 19.6 m/s^2
h = 35 m.
Solve for V.
26.19
To find the speed of the rock after it has fallen a vertical distance of 35m, we can use the principles of energy conservation.
First, we need to calculate the potential energy (PE) that the rock has at the top of the cliff. The potential energy formula is given by PE = m * g * h, where m is the mass of the rock, g is the acceleration due to gravity, and h is the vertical distance.
PE = 15 kg * 9.8 m/s^2 * 35 m
PE = 5,136 J (joules)
Next, we need to calculate the final kinetic energy (KE) of the rock. The initial kinetic energy (KE_initial) is equal to the initial potential energy, as there is no air resistance and no other forces acting on the rock.
KE_initial = PE = 5,136 J
Finally, using the equation for kinetic energy, KE = (1/2) * m * v^2, we can find the final velocity (v) of the rock.
KE = (1/2) * m * v^2
5,136 J = (1/2) * 15 kg * v^2
Simplifying the equation:
v^2 = (2 * 5,136 J) / 15 kg
v^2 = 342.4 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 18.5 m/s
So, the speed of the rock after it has fallen a vertical distance of 35m is approximately 18.5 m/s.