Question

What is the hang time of a person who can jump a vertical distance of $0.6 m$?

Answer 1

Hint: We have to find the time for how much that person remains in air i.e., hang time. Person jumps freely in this case; acceleration is equal to the acceleration due to gravity and it jumps with zero initial velocity. With the help of a second equation of motion, we can find the hang time.

Complete step-by-step solution:
Given: distance travelled by person, $s = 0.6 m$.
Initial velocity, $u = 0$.
We have to find the time taken by a person. We will use the second equation of motion.
$s = ut + \dfrac{1}{2} a t^{2}$
Where, s is the displacement of the body (in this case it is height)
a is the acceleration of the person and here $a=g$.
t is the time taken to reach the distance. (Hang time)
Now put the values and find the value of hang time.
$0.6 = 0t + \dfrac{1}{2} \times 9.8 \times t^{2}$
$\implies t = \sqrt{\dfrac{0.6 \times 2}{9.8}}$
$\implies t = 0.3499 s$
Hence, the hang time of a person is $0.3499 s$.
Additional Information: - Hang time is the time that a person takes to reach that height from ground, but when a person comes back to the ground then its down time can also be calculated i.e., same as hang time. Total time is the sum of time going up and time coming down. So, we can say that-
$\textit{Total time} = 2 \times \textit{hang time}$

Note: In the start, the person is at the ground (at rest) so, we take initial velocity as zero. It jumps upward without any external force, so acceleration becomes equal to the acceleration due to gravity. We take acceleration positively as time cannot be a complex number, it should be real.