Question

A body freely falls from a certain height on to the ground in a time $t$. During the first one third of the time interval it gains a kinetic energy $\Delta k_1$ and during the last one-third of the interval, it gains a kinetic energy $\Delta k_2$. The ratio $\Delta k_1:\Delta k_2$ is:

• A. $1:1$
• B. $1:3$
• C. $1:4$
• D. $1:5$

Answer 1

Answer: (D)

$1:5$

$S=\frac{1}{2}g{t}^2$

For the first one third of time:

$\Delta K_1\propto (\frac{t}{3}{)}^2$

For the last one third of time:

$\Delta K_2\propto (t^2-(\frac{2t}{3}{)}^2)$

$\frac{\Delta K_1}{\Delta K_2}=\frac{(\frac{t}{3}{)}^2}{(t^2-(\frac{2t}{3}{)}^2)}=\frac{1}{5}$