Question

Two projectile A and B thrown with speed in ratio 1:√2 acquiring same height if A is thrown at an angle 45° with the horizontal the angle of projection of B will be?

Answer 1

The angle of projection of B should be 30°.

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

=> Va²  sin² 45 / 2g = Vb² sin²Ф / 2g

=> sin²Ф = (Va²/Vb²) sin² 45

=  1/4

=> sinФ = 1/2

=> Ф = 30°

Hence the angle of projection of B should be 30°.

Answer 2

Answer:

using the formula,

H = v^2 sin^2 ¤ / 2g

with same height and velocity ,

the angle of projection at B will be 30°.