Question

a projectile is projected at an angle of 15 degree to the horizontal with speed v . If another projectile is projected with the same speed then at what angle with the horizontal it must be projected so as to have the same range

Answer 1

it is projected with the angle 90-15 = 75 degree...because to have same range..angle of projection should be complementry...

Answer 2

Answer:

The correct answer is 75 degrees.

Explanation:

There are still two values of $\theta$ (angle of projection for which the projectile has the exact range Those two angles are $\theta$ and $(90-\theta)$.

we have

range = $\frac{2 \cdot u^{2} \cdot \sin \theta \cdot \cos \theta}{g}$

now let the angle be $(90-\theta)$

then range $=\frac{2 \cdot u^{2} \cdot \sin (90-\theta) \cdot \cos (90-\theta)}{g}$

simplifying we get

$=\frac{2 \cdot u^{2} \cdot \sin \theta \cdot \cos \theta}{g}$

we see that the range exists the identical irrespective of the different times of flight'

so, in the question given one angle 15 degrees

then the other angle exists $(90-\theta) ;(90-15) ;=75$ degree.

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