# Civil Engineering - Applied Mechanics - Discussion

### Discussion :: Applied Mechanics - Section 2 (Q.No.5)

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.

#### Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.

#### Current Affairs 2021

Interview Questions and Answers