A particle is projected with a velocity u in a direction making an angle θ with
the horizontal. Find (i) the maximum height (ii) time of flight.
(b) Prove that the time of flight T and the horizontal range R of a projectile are
connected by the equation g T 2 = 2 R tan θ.
Answers
Answer:
11th
Physics
Motion in a Plane
Projectile Motion
A projectile is thrown at a...
Physics
A projectile is thrown at an angle θ from the horizontal with velocity 'u' under the gravitation field of the earth. Derive expressions for its:
a)Time of its flight
b)Height
c)Horizontal Range
Easy
Answer
Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .
The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.
Now,
(a). The total time of flight is
Resultant displacement is zero in Vertical direction.
Therefore, by using equation of motion
s=ut−
2
1
gt
2
gt=2sinθ
t=
g
2sinθ
(b). The horizontal range is
Horizontal range OA = horizontal component of velocity × total flight time
R=ucosθ×
g
2usinθ
R=
g
u
2
sin2θ
(c). The maximum height is
It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.
By using equation of motion
v
2
=u
2
−2as
0=u
2
sin
2
θ−2gH
H=
2g
u
2
sin
2
θ
Explanation:
I hope it will be help full for you