Question

If Michael Jordan has a vertical leap of $1.29m$, then what is his hang time (total time to move upwards to the peak and then return to the ground)?

• A. $2.03s$
  
• B. $1.03s$
  
• C. $0.03s$
  
• D. $1.0s$
  

Answer 1

Answer: (B)

$1.03s$

Given, acceleration $a=-g=-9.8m/s^2$ (minus sign for motion against gravity); final velocity (top most point) is $v=0m/s$

Vertical distance traveled $d=1.29m$

If $u$ be the initial velocity, using formula $v^2-u^2=2ad$

$0^2-u^2=2(-9.8)\left(1.29\right)$

$⟹$ $u=5.03m/s$

If $t$ be the time to peak.

Using $v=u+at$

$0=5.03-\left(9.8\right)t$

We get $t=5.03/9.8=0.513s$

So, the hang time will be double of peak time i.e. $t_{hang}=2\times 0.513=1.03s$