Q1: The angle of projection for a range is equal to the distance through which the particle would have fallen in order to acquire a velocity equal to the velocity of projection, will be

ANS:D - 75°

Let a body is projected with speed u m/s at an angle α with horizontal ( or angle of projection is α).

then, the horizontal range of the body, R = u^2sin2α/g ---> (1)

a/c to question,

range = distance covered by a body in the vertical direction in which the body acquires velocity equals to velocity of projection.

i.e., R = h ---> (2)

the initial velocity of the body along vertical direction = 0.

Final velocity, v = u.

Then using formula, v^2 = u^2 + 2as.

or, u^2 = 0 + 2gh.

or, h = u^2/2g ---> (3)

from equations (1), (2) and (3),

u^2sin2α/g = u^2/2g

or, sin2α = 1/2 = sin30° or sin150°
or, 2α = 30° ; or 150°
or, α = 15° or 75°

Hence, angle of projection : 15° or 75°