Question

if michael jordan has a vertical leap of 1.29m,then what is his takeoff speed and his hang time(total time to move upwards to the peak and then return to the ground).

Answer 1

Let the initial velocity be ‘u’.

Acceleration due to gravity, g = -9.8 m/s2 [downward]

Let the time of flight be ‘t’.

So, using

v2 = u2 + 2as

=> 0 = u2 + 2(-9.8)(1.29)

=> u = 5.03 m/s

Now,

s = ut + ½ at2

=> 0 = 5.03t – 0.5 × 9.8 × t2

=> t = 1.03 s

Answer 2

Answer:

The takeoff speed and his hang time are 5.08m/s and 1.016 seconds respectively.

Explanation:

The takeoff speed is given as,

$v=\sqrt{2gh}$           (1)

Where,

v=takeoff speed

g=acceleration due to gravity=10m/s²

h=height from which the vertical leap is taken

The hang time (T) is given as,

$T=\sqrt{\frac{2h}{g} }$         (2)

From the question we have,

h=1.29m

By substituting the values of acceleration due to gravity and height of vertical leap in equation (1) we get;

$v=\sqrt{2\times 10\times 1.29}$

$v=\sqrt{25.8} =5.08m/s$       (3)

By substituting the required values in equation (2) we get;

$T=\sqrt{\frac{2\times 1.29}{10} }$

$T=\sqrt{0.258}$

$T=0.508 sec$        (4)

And the total time is given as,

$2\times 0.508=1.016sec$     (5)

Hence, the takeoff speed and his hang time are 5.08m/s and 1.016 seconds respectively.