Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : 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
Discussion:
7 comments Page 1 of 1.
Suresh reddy said:
6 years ago
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.
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.
(1)
Deep said:
7 years ago
Shouldn't it be like v= √(v * 2ghcotθ)?
As the horizontal distance for 60° and 100 m vertical is 100cot60.
As the horizontal distance for 60° and 100 m vertical is 100cot60.
Abhay said:
10 years ago
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.
v = sqrt(u^2 - 2gh).
v = sqrt(100.3^2 - 2*9.81*100).
v = 90 m/sec.
Utsav said:
9 years ago
It is wrong.
The correct one is:
v = sqrt((u * sin60)^2 - 2gh).
v = 75.
The correct one is:
v = sqrt((u * sin60)^2 - 2gh).
v = 75.
Ahmad Mujtaba said:
5 years ago
v = √(v^2 - 2gh),
v = √(100.3^2 - 2*9.8*100).
v = 90 m/sec.
v = √(100.3^2 - 2*9.8*100).
v = 90 m/sec.
Ravi Mehta said:
9 years ago
As per me, option B is correct.
Sachin Singh said:
9 years ago
How is this right?
V = 75.
V = 75.
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