Question

A particle is projected upwards from level ground at an angle with horizontal so that its range is twice the maximum height reached by it. the angle made by its velocity vector with the horizontal at the horizontal distance R/4 from the point of projection is

30°

60°

45°

15°​

Answer 1

Answer:

60° is the right answers

Answer 2

Given info : A particle is projected upwards from level ground at an angle with horizontal so that its range is twice the maximum height reached by it.

To find : the angle made by its velocity vector with the horizontal at the horizontal distance R/4 from the point of projection is ..

solution : let particle is projected at angle Ф with horizontal with speed u.

horizontal range = u² sin2Ф/g ...(1)

maximun height = u²sin²Ф/2g

a/c to question,

    horizontal range = 2 × maximum height

⇒ u²sin2Ф/g = 2 × u²sin²Ф/2g

⇒ 2sinФ cosФ = sin²Ф [ we know, sin2Ф = 2sinФ cosФ ]

⇒ tanФ = 2

when particle moves R/4 distance in horizontal direction.

velocity in horizontal direction remains constant and that is ucosФ

but the velocity in vertical direction changes which is usinФ - gt

x = R/4 = ucosФ t

⇒ t = R/(4ucosФ) = usinФ/2g [ from eq (1) }

so, vertical component of velocity = usinФ - g (usinФ/2g) = usinФ/2g

so the angle made by velocity vector , tanα = $\frac{v_y}{v_x}$ = tanФ/2 = 2/2 = 1 = tan45°

so the angle made by the velocity vector at the horizontal direction R/4 from the point of projection is 45°