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

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Answer 1 · 2017-09-30T10:07:23

0.3499 seconds

The person jumps a height of 0.6 m .

Remember the kinematic equation ,

$S = u t + \frac{1}{2} a t^{2}$ $S=ut+1/2at^2$

where ,

S is the displacement of the body (in this case it is height from which the person jumps )

u is the initially velocity . Here it is zero because the person jumps freely under the action of gravity and initially he is at rest .

a is the acceleration of the person .

t is the time taken to reach the distance . (Hang time)

Now substitute the values and find the value of hang time .

$0.6 = 0 \times t + \frac{1}{2} g \cdot t^{2}$ $0.6=0xxt+1/2g*t^2$ .

g is the acceleration of the person . It is the acceleration due to gravity whose value is $9.8 m s^{- 2}$ $9.8 ms^-2$ .

$0.6 = \frac{1}{2} \times 9.8 \times t^{2}$ $0.6=1/2xx9.8xxt^2$

$\frac{0.6 \times 2}{9.8} = t^{2}$ $(0.6xx2)/9.8=t^2$

$t = \sqrt{\frac{1.2}{9.8}}$ $t=sqrt(1.2/9.8)$

$t = 0.3499 s$ $t=0.3499 s$