Question

An athlete of mass $87.4kg$ jumps off the ground. The peak of his jump is $2.2m$. Given that the athlete is completely at the peak, what was his takeoff velocity?

Answer 1

We can find out the potential energy of the athlete when he is at the peak, which will be its maximum potential energy. From the conservation of mechanical energy, the maximum kinetic energy of the athlete will be equal to his maximum potential energy. From there we can find out the required takeoff velocity.

Formula used: The formulae used for solving this question are given by
$U = mgh$, here $U$ is the potential energy of a particle of mass $m$ at a height of $h$, and $g$ is the acceleration due to gravity.
$K = \dfrac{1}{2}m{v^2}$, here $K$ is the kinetic energy of a particle of mass $m$ which is moving with a velocity of $v$.

Complete step by step answer
When the athlete reaches the peak, his potential energy becomes maximum. We know that the potential energy is given by
$U = mgh$
According to the question, the maximum height of the athlete is ${h_{\max }} = 2.2m$, his mass is $87.4kg$. So the maximum potential energy is given by
${U_{\max }} = mg{h_{\max }}$
$\Rightarrow {U_{\max }} = 87.4 \times 9.8 \times 2.2$
On solving we get
${U_{\max }} = 1884.344J$ (1)
Now, at the instant when the athlete jumps off the ground, his potential energy is equal to zero. Therefore is energy is entirely kinetic. So at this instant his kinetic energy is maximum. Since only the gravitational force acts on the athlete during his motion, which is a conservative force, so his total mechanical energy is conserved. Therefore the maximum kinetic energy equals to the maximum potential energy. Let $v$ be the takeoff velocity of the athlete.
So the maximum kinetic energy is
${K_{\max }} = \dfrac{1}{2}m{v^2}$ (2)
Since ${K_{\max }} = {U_{\max }}$, so from (1) and (2) we have
$\dfrac{1}{2}m{v^2} = 1884.344$
$\Rightarrow {v^2} = \dfrac{{2 \times 1884.344}}{m}$
Substituting $m = 87.4kg$ we get
${v^2} = \dfrac{{2 \times 1884.344}}{{87.4}}$
$\Rightarrow {v^2} = 43.12$
Finally, on taking square root on both the sides, we get
$v = 6.57m/s$
Hence, the takeoff velocity of the athlete is equal to $6.57m/s$.

Note
This question could be attempted even without using the value of the mass of the athlete. When the maximum kinetic energy is equated with the maximum potential energy, then the mass gets cancelled from both the sides. So there is no use of the value of mass in solving this question.