Civil Engineering - Applied Mechanics - Discussion

5. 

A projectile is fired with a velocity of 100.3 m/sec. at an elevation of 60°. The velocity attained by the projectile when it is moving at a height of 100 m, is

[A]. 70 m/sec.
[B]. 75 m/sec.
[C]. 80 m/sec.
[D]. 85 m/sec.
[E]. 90 m/sec.

Answer: Option E

Explanation:

No answer description available for this question.

Abhay said: (Jan 25, 2016)  
The velocity of a projectile a height h is given by:

v = sqrt(u^2 - 2gh).
v = sqrt(100.3^2 - 2*9.81*100).
v = 90 m/sec.

Utsav said: (Jul 21, 2016)  
It is wrong.
The correct one is:
v = sqrt((u * sin60)^2 - 2gh).
v = 75.

Sachin Singh said: (Dec 31, 2016)  
How is this right?

V = 75.

Ravi Mehta said: (Jan 8, 2017)  
As per me, option B is correct.

Deep said: (Dec 4, 2018)  
Shouldn't it be like v= √(v * 2ghcotθ)?
As the horizontal distance for 60° and 100 m vertical is 100cot60.

Suresh Reddy said: (Mar 22, 2019)  
As per when the moves upward it lose their velocity at the rate of(9.8m/S)/S.
In vertical direction velocity ---u^2sin^2(60).
(V@vertical)^2 = u2sin^2(60)-2gh,
(V@horizontal)^2 =u^2cos^2(60)+2gh,
(Results v)^2 = v1^2+v2^2-2v1 * v2cos60.
As per me, answer 70(+-)1m/S.
Option A.

Ahmad Mujtaba said: (Feb 10, 2021)  
v = √(v^2 - 2gh),
v = √(100.3^2 - 2*9.8*100).
v = 90 m/sec.

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