Question

A coin is dropped from a height of 80 m. How fast will it fall, just before it hits the ground? [Take g = 10 m/s2 , Hint: For a falling body, the total PE at the greatest height will be equal to the K.E., just before it touches the ground]​

Answer 1

The speed at which the coin falls just before it hits the ground is 40 m/s.

Step-by-step explanation:

The speed at which a coin falls just before it hits the ground can be found by equating its potential energy at the greatest height to its kinetic energy just before it touches the ground. The potential energy (PE) is given by the formula PE = mgh, where m is the mass of the coin, g is the acceleration due to gravity, and h is the height. In this case, the height is 80 m.

The kinetic energy (KE) is given by the formula KE = 1/2 mv², where v is the speed of the coin. Equating PE and KE, we get mgh = 1/2 mv². Simplifying the equation, we get v = √(2gh). Substituting the given values,

we get v = √(2 * 10 * 80) = √(1600) = 40 m/s.