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

##### 2 Answers
Dec 4, 2015

$0.32 \text{s}$

#### Explanation:

The equation of motion says:

$v = u + a t$

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

$0 = {u}_{y} - \text{g} t$

$u \sin \theta = {u}_{y}$ which is the vertical component of his initial velocity $u$.

$\therefore 0 = u \sin \theta - \text{g} t$

$\therefore t = \frac{u \sin \theta}{g}$

Since $\theta = 23$ this becomes:

$t = \frac{4 \sin 23}{9.8} = 0.159 \text{s}$

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"

Dec 4, 2015

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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