Question

Michael Jordan has a vertical leap of $1.25m$. What is his takeoff speed and for how much time he will be in the air?
Take $a=10m{s}^{-2}$ downwards.

• A. $\text{u = 10}m{s}^{-1},\text{t = 1 s}$
• B. $\text{u = 15}m{s}^{-1},\text{t = 1.5 s}$
• C. $\text{u = 5}m{s}^{-1},\text{t = 1 s}$
• D. $\text{u = 10}m{s}^{-1},\text{t = 0.5 s}$

Answer 1

The correct option is C $\text{u = 5}m{s}^{-1},\text{t = 1 s}$

Given:
Distance travelled one side $s=1.25m$
Acceleration, $a=-10m{s}^{-2}$ (his motion is against gravity)
Final velocity, $v=0m{s}^{-1}$ ( since he is at rest after takeoff)

Let '$u$' be the initial velocity and '$t$' be the time taken.
From the third equation of motion, $v^2=u^2+2as$,
$0=u^2-2\times 10\times 1.25$
$\Rightarrow$ $u=5m{s}^{-1}$

From the first equation of motion, $v=u+at$,
$0=5-10\times t$
$\Rightarrow t=\frac{5}{10}=0.5s$

Since the time taken for going up and coming down will be equal. Total time for which he remained in air is $1s$.