$(a)$ Derive an expression for the kinetic energy of a body having mass $m$ and moving with a velocity $v$.
$(b)$ When the velocity of a body is increased $5$ times,what is the change in its kinetic energy?
$(c)$ Two masses $m$ and $2m$ are dropped from heights $h$ and $2h$. On reaching the ground,which will have greater kinetic energy and why?

(N/A) Kinetic energy $(KE)$ is the work done by a force $F$ to accelerate a body of mass $m$ from rest $(u=0)$ to velocity $v$ over a distance $s$. Using $v^2 - u^2 = 2as$,we get $s = v^2 / (2a)$. Since $F = ma$,the work done $W = F \times s = ma \times (v^2 / 2a) = 1/2 mv^2$. Thus,$KE = 1/2 mv^2$.
$(b)$ Since $KE \propto v^2$,if the velocity becomes $5v$,the new kinetic energy $KE' = 1/2 m(5v)^2 = 25 \times (1/2 mv^2) = 25 \times KE$. The kinetic energy increases by a factor of $25$.
$(c)$ The kinetic energy of a falling body at the ground is equal to its initial potential energy $(PE = mgh)$. For mass $m$ at height $h$,$PE_1 = mgh$. For mass $2m$ at height $2h$,$PE_2 = (2m)g(2h) = 4mgh$. Since $PE_2 > PE_1$,the body with mass $2m$ will have greater kinetic energy.