# Civil Engineering - Applied Mechanics - Discussion

Discussion Forum : Applied Mechanics - Section 1 (Q.No. 50)
50.
If a particle is projected inside a horizontal tunnel which is 554 cm high with a velocity of 60 m per sec, the angle of projection for maximum range, is
10°
11°
12°.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Muhammad Umar said:   6 months ago
The answer is required for maximum range, not for maximum height.

For the Maximum range (149.413 m) the angle must be 12 degrees.

Range values for other given angles are less than the range occupied if the angle of trajectory is 12 degrees.

Greeshma said:   4 years ago
This is a projectile motion how can we use this equation. The equation of max height of the projectile is (v^2sin^2angle)/(2g) right?

MANOJ KUMAR said:   5 years ago
H=u^2 sin^2angle/2g.

5.54=60^2sin^2angle/(2*9.81),
after solving this angle is 10.00650.
So, the answer is 10.
(2)

Pushkar jha said:   4 years ago
By putting formula for maximum height. We can find the angle.

Kiran said:   7 years ago
By using this;

H=u^2 sin^angle/2g we can get the answer.

Mayur said:   5 years ago
Thank you all for explaining this.

Mani mala said:   6 years ago
Please explain it detaily.

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