Question

Question 14 5 pts A basketball is dropped from rest from a height h. If the height from which the ball is released is doubled and the experiment repeated, the change in kinetic energy of the ball will be that of the original ball. Half One quarter Quadruple Double

Answer 1

When it is at height h,

potential energy is   

kinetic energy is  

When it reaches the ground,

potential energy is   

kinetic energy is  $K = \frac{1}{2} mv^2$

where v is velocity at the end

Change in Kinetic energy = Change in potential energy

mu 0= mgh - 0 2 1 5 mu? = mgh V = V2gh

Thus if the height is doubled, then velocity becomes

$\\v' = \sqrt{2g(2h)} \\\\v' = \sqrt{2}\sqrt{2gh} \\\\v' = \sqrt{2}\ v$

change in kinetic energy in the first case is,

$\\\Delta K_1 = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$

change in kinetic energy in the second case is,

$\\\Delta K_2 = \frac{1}{2}m(v')^2 - 0 \\\\\Delta K_2= \frac{1}{2}m(\sqrt{2}v)^2 \\\\\Delta K_2 = 2 * \frac{1}{2}mv^2 \\\\\Delta K_2 = 2 * \Delta K_1$

therefore, change in kinetic energy of the ball will be double that of the original ball.