A ball thrown horizontally from a height of 20 m and hits the ground with a speed that is three times its initial speed. What was the initial speed?

Answer 1

This problem is best worked with conservation of energy. In this problem, when there is no friction or external forces, Mechanical energy is conserved.

Let's assume that the initial velocity is v and therefore that the final velocity is 3v.

The initial total energy is potential energy plus kinetic = mg(20 meters) + 1/2 mv²

The final energy, once the ball is right above the ground is all kinetic = 1/2 m(3v)²

Let's set those two equal

m(9.8)(20) + 1/2mv² = 1/2 m* 9 * v²

we can divide through by m and rearrange the equation to get the following

196 = (8/2) V²

^{V = 7 m/sec}

^{Let's check the initial ME = m(9.8)(20) + 1/2(m)(49) = 220.5 m}

the final ME = 1/2(m)*(21)^2 = 220.5m

Since they match, our solution is correct