Question

An athlete jumps with the speed of 4 m/s at an angle of 23°. How long does the athlete stay in the air?

Answer 1

`0.32"s"`

Explanation:

The equation of motion says:

`v=u+at`

Considering only the vertical component of his jump we can get the time it takes to reach his maximum height from:

`0=u_y-"g"t`

`usintheta=u_y` which is the vertical component of his initial velocity `u`.

`:.0=usintheta-"g"t`

`:.t=(usintheta)/g`

Since `theta=23` this becomes:

`t=(4sin23)/9.8=0.159"s"`

The 2nd part of his journey as he falls to earth is a mirror image of the first part so:

`t_("tot")=2xx0.159=0.32"s"`

Answer 2

I found `0.32` seconds but please check my maths!

Explanation:

I considered the accelerated (downwards) vertical component of the movement only.
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