Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 14)
14.
The time of flight (t) of a projectile on an upward inclined plane is(where u = Velocity of projection, α = Angle of projection, and β = Inclination of the plane with the horizontal.)
Discussion:
13 comments Page 1 of 2.
Harendra said:
8 years ago
S= ut + 0.5 a (t)^2.
here S=0, since range is horizontal distance covered until the particle reached the its initial point.
overall angle of projection = (α-β).
accelaration (a) = g*cosβ , since inclination of the plane is present.
u = intial velocity in x direction = usin(α-β).
Substitute all the values in the equation. You will get the answer.
here S=0, since range is horizontal distance covered until the particle reached the its initial point.
overall angle of projection = (α-β).
accelaration (a) = g*cosβ , since inclination of the plane is present.
u = intial velocity in x direction = usin(α-β).
Substitute all the values in the equation. You will get the answer.
(4)
Kumara swamy said:
9 years ago
Rotate your coordinate axis by beta degrees and resolve your "g" component in your new axis.
Your displacement in y-axis will be given as.
y = usin (a-b) - 0.5cos (b) t^2.
After a time of flight T displacement along y-axis will be zero places y = 0 in the equation, you will get the answer.
Your displacement in y-axis will be given as.
y = usin (a-b) - 0.5cos (b) t^2.
After a time of flight T displacement along y-axis will be zero places y = 0 in the equation, you will get the answer.
Kiran P. said:
9 years ago
The total time taken by a projectile to reach the maximum height and to return back to ground is known as the time of flight.
Mahesh said:
1 decade ago
When inclination is increased automatically the time of flight increases there by the denominator should be minimum.
Arun kumar dewangan said:
1 decade ago
This answer will be possible only when projectile motion normal to the inclined plane, where displacement is zero.
Suresh s said:
9 years ago
@K.Kisku.
While throwing an object, the time taken by the object to reach rest or destination.
While throwing an object, the time taken by the object to reach rest or destination.
KEVINES said:
8 years ago
Here the angle of projection is also from horizontal.
M Saikrishna said:
8 years ago
According to me, the answer is (2ucos(a))/(gcos(b)).
Rajeev said:
7 years ago
@Harendra.
Here, we can use a quadratic equation.
Here, we can use a quadratic equation.
Vishal said:
1 decade ago
There's no range (distance) given in the options.
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