Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : Applied Mechanics - Section 4 (Q.No. 37)
37.
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
30°
45°
60°
75°.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
3 comments Page 1 of 1.

Muhammad Hashim BZU Multan said:   1 year ago
No, the right answer should be 60 degrees.

76 degrees will be right when height = Range.

When initial velocity (Projection) = Final velocity (Just before striking the ground) the angle of Projection will be 60 degrees.

Naveed Ahmad said:   4 years ago
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°

G Naveen said:   6 years ago
Please explain the answer.

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