Physics 2011 JAMB Past Questions - Question 2: A carpenter on top of roof 20.0m high dropped a hammer of mass 1.5 kg and it fell freely to the ground. The kinetic energy of the hammer just before hitting the ground is

Correct Answer

A

Explanation

K.E = P.E = Mgh = 1.5 x 10 x 20 = 300JTo find the kinetic energy of the hammer just before hitting the ground, you can use the following formula for kinetic energy (KE):

KE = 0.5 * m * v^2

Where:
KE = Kinetic energy
m = Mass of the hammer (1.5 kg)
v = Velocity of the hammer just before hitting the ground

First, you need to find the velocity of the hammer just before it hits the ground. You can use the equations of motion to do that. The hammer is in free fall, so you can use the equation:

h = (1/2) * g * t^2

Where:
h = Height (20.0 m)
g = Acceleration due to gravity (approximately 9.81 m/s^2 on the surface of the Earth)
t = Time

You can solve this equation for time, t:

t = sqrt((2 * h) / g)

Substitute the values:

t = sqrt((2 * 20.0 m) / 9.81 m/s^2) ≈ 2.02 seconds

Now that you have the time it takes for the hammer to fall, you can find the final velocity using the formula for velocity during free fall:

v = g * t

v = 9.81 m/s^2 * 2.02 s ≈ 19.84 m/s

Now that you have the velocity, you can calculate the kinetic energy:

KE = 0.5 * 1.5 kg * (19.84 m/s)^2

KE ≈ 281.76 joules

So, the kinetic energy of the hammer just before hitting the ground is approximately 281.76 joules.