Question

a certain athlete consistently throws a javelin at a speed 25m/s.what is her best distance?on one occasion the athlete released the javelin poorly,and achieved only one half of his distance.at what elevation angle did she release the javelin

Answer 1

Question#19157

a certain athlete consistently throws a javelin at a speed 25m/s. what is her best distance? on one occasion the athlete released the javelin poorly, and achieved only one half of his distance. at what elevation angle did she release the javelin

Solution:

The best distance is if elevation angle equal to 45°

The distance is:

$S = v_{x}t$ were $t$-time of flight of a javelin, $v_{x}$- the horizontal component of velocity.

$v_{x} = v\cos \alpha$, were $\alpha$-the elevatin angle

Such as the javelin move with by gravity force:

$v_{y} = gt$, $t = \frac{v_{y}}{g}$ were $v_{y}$-vertical component of velocity, $g$-the gravity acceleration.

$$v_{y} = v \sin \alpha$$ $$t = 2 \frac{v \sin \alpha}{g}$$ $$S = 2 \frac{v^{2} \sin \alpha \cos \alpha}{g} = \frac{v^{2} \sin 2\alpha}{g}$$ $$S = \frac{25^{2} \sin (2 \times 45{}^{\circ})}{9.8} = 63.78 \, \text{m}$$

If the distance is a half of S,

such as $\sin (2 \times 45{}^{\circ}) = \sin 90{}^{\circ} = 1$, the angle will be:

$$\sin \alpha = 0.5$$ $$\alpha = \arcsin 0.5 = 30{}^{\circ}$$

Answer: distance is: 63.78 m, angle is: 30°