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A prankster has dropped a 173 kg water balloon from a cargo plane targeting a high-rise building that is 61.7 m tall. What will be the impact velocity of the water balloon when it hits the ground?

Answer :

Final answer:

The impact velocity of the water balloon when it hits the ground, ignoring air resistance, will be about 35 m/s. This was calculated using the physics principle of conservation of energy in free fall.

Explanation:

This problem involving the falling water balloon can be solved using the physics of free fall and the principle of conservation of energy. We know that gravitational potential energy at the beginning is transformed into kinetic energy at the end of the fall. That gives us the equation for kinetic energy: E_k = 1/2 m v^2.

However, we want to figure out the final velocity of the balloon, v. To solve for this, we know that the initial gravitational potential energy of the balloon is mgh, where m is the mass of the balloon (173kg), g is the gravitational acceleration (~9.81 m/s²), and h is the height from which the balloon is dropped (61.7m).

By equating these two forms of energy (since energy is conserved), we get the equation 1/2 m v^2 = m g h. Solving for v, we find that the speed of the balloon when it hits the ground, ignoring factors like air resistance, will be about 35 m/s.

Learn more about Free Fall here:

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