# A projectile is projected at an angle of 15° 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 yo have the same range?

Jun 22, 2018

$\theta = 75 \mathrm{de} g r e e$

#### Explanation:

Note :- There are always two values of $\theta$(angle of projection for which the projectile has the same range .Those two angles are
$\theta \mathmr{and} \left(90 - \theta\right)$

we have range=
$\frac{2 \cdot {u}^{2} \cdot \sin \theta \cdot \cos \theta}{g}$
now let the angle be $\left(90 - \theta\right)$
then range= $\frac{2 \cdot {u}^{2} \cdot \sin \left(90 - \theta\right) \cdot \cos \left(90 - \theta\right)}{g}$
simplifying we get=$\frac{2 \cdot {u}^{2} \cdot \sin \theta \cdot \cos \theta}{g}$
we see that the range is same irrespective of the different time of flight'
so in the question given one angle $15 \mathrm{de} g r e e$
then the other angle is (90-theta);(90-15);=75degree
Note:- in this it was given that both the projectile have the same speed hence we can think both of them as a same projectile with different angles of projection.Never use this when the speeds are different.