Question

Two projectile A and B thrown with speed in the ratio acquired the same heights.If A is thrown at an angle of projection of B will be.

Answer 1

From the projectile motion we know that the height attained by a projectile is given by :

$H=\frac{v^2{sin}^2\alpha }{2g}$

where v = speed

α = angle

Now both A and B acquire same height, hence

$\frac{{Va}^2{sin}^{2}45}{2g}=\frac{{Vb}^2{sin}^2\theta }{2g}$

${sin}^2\theta =\frac{{Va}^2}{{Vb}^2}{sin}^{2}45$

$sin\theta =\frac{1}{2}$

$\theta ={30}^o$