Question

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

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

Answer 1

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

Given, distance travelled one side s = 1.25 m
and 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 1 second.