Question

An athlete accelerates to 10 m/sex and then jumps at a 25° angle. How far Will he jumps?

Answer 1

Question #71525, Physics / Mechanics | Relativity

An athlete accelerates to $10\,\mathrm{m}/\mathrm{sex}$ and then jumps at a $25{}^{\circ}$ angle. How far Will he jumps?

Answer: Draw -

$$v_0 = 10 \frac{m}{s}$$ $$\alpha = 25{}^{\circ}$$

Need to find: $l - ?$

$$l = x = v \cdot t = v_0 \cdot \cos \alpha \cdot t$$

y - direction:

$$y = y_0 + v_0 \cdot t + \frac{g \cdot t^2}{2},\ 0 = 0 + v_0 \cdot \sin \alpha \cdot t - \frac{g \cdot t^2}{2}$$

Then - $v_0 \cdot \sin \alpha = \frac{g \cdot t}{2},\ t = \frac{2 \cdot v_0 \cdot \sin \alpha}{g}$ - the flight time of the body to fall to the ground.

$$l = v_0 \cdot \cos \alpha \cdot \frac{2 \cdot v_0 \cdot \sin \alpha}{g} = \frac{v_0^2 \cdot (2 \cdot \cos \alpha \cdot \sin \alpha)}{g}$$

From mathematics it is known that $-2 \cdot \cos \alpha \cdot \sin \alpha = \sin 2\alpha$, then

$$l = \frac{v_0^2 \cdot \sin 2\alpha}{g}$$

let's calculate, $l = \frac{10^2 \frac{m^2}{s^2} \cdot \sin 2 \cdot 25}{9.8 \frac{m}{s^2}} \approx 7.81\, \mathrm{m}$

He jumps on 7,81 meters.

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