# 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

Discussion:

7 comments Page 1 of 1.
MANOJ KUMAR said:
6 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.

5.54=60^2sin^2angle/(2*9.81),

after solving this angle is 10.00650.

So, the answer is 10.

(2)

Kiran said:
8 years ago

By using this;

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

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

Mani mala said:
6 years ago

Please explain it detaily.

Mayur said:
6 years ago

Thank you all for explaining this.

Pushkar jha said:
5 years ago

By putting formula for maximum height. We can find the angle.

Greeshma said:
5 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?

Muhammad Umar said:
1 year 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.

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.

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