A particle is projected from horizontal ground at angle 'theta' with speed 'u'. In same plane of motion a horizontal acceleration 'a' exists so that projected particle returns back to point of projection. Find time of flight.
Answers
Answer:
Take X,Y axes as shown in fugure. Suppose that the particle strikes the plane at a point P with coordinates (x,y). Consider the motion between A and P.
Now,
Motion in x direction =u
Acceleration a=0
And x=ut
Now
Motion in y direction
Initial velocity u=0
Acceleration a=g
So,
y=ut+
2
1
gt
2
=
2
1
gt
2
y=
2
!
gt
2
Now eliminate t from x and y respectively, we get
y=xtanθ..........(1)
Thus,
xtanθ=
2u
2
gt
2
On solving for x get two values i.e.
x=0 and x=
g
2u
2
tanθ
Therefore the point is corresponding to x=
g
2u
2
tanθ
So put this value of x in equation (1) we get,
y=xtanθ=
g
2u
2
tan
2
θ
Then,
The distance is OP=
x
2
+y
2
=
g
2u
2
tanθ
1+tanθ
=
g
2u
2
tanθsecθ
Answer:
The time of flight for a particle projected from the horizontal ground at angle θ with speed u is 2u sin θ/ g.
Explanation:
- A distance travelled by an item, particle, or wave across a medium is measured in terms of the time of flight.
- The velocity or length of the journey can then be determined using this information, as well as the characteristics of the particle or medium.
- Direct or indirect detection of the moving object is both possible.
- Time of Flight is influenced by the projectile's angle and beginning velocity magnitude.
- It is given by the formula- T = 2u sin θ/ g where u is velocity magnitude, g is the acceleration due to gravity, θ is the angle of the projectile and T is the time of flight.
Thus, the time of flight when a particle is projected from the horizontal ground at angle θ with speed u is 2u sin θ/ g.
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