Question

A track star in the broad jump goes into the jump at 12 m/s and launches himself at 20° above the horizontal. How long is he in the air before returning to Earth?

Answer 1

Explanation:

It is given that,

Initial speed of the broad jump, u = 12 m/s

It is launched at an angle of 20 degrees above the horizontal. Let t is the time for which the track star i in the air before returning to Earth. The motion of the track star in the broad jump can be treat as the projectile motion. The time of flight of the projectile is given by :

$t=\dfrac{2u\ sin\theta}{g}$

Putting all the values in above equation as :

$t=\dfrac{2\times 12\times \ sin(20)}{9.8}$

t = 0.837 seconds

So, the time for which the track star is in air is 0.837 seconds. Hence, this is the required solution.