Hey Frndz
1)DERIVE EQUATION OF TRAJECTORY FOR HORIZONTAL PROJECTILE.
2)DERIVE THE EXPRESSION FOR TIME OF FLIGHT AND HORIZONTAL RANGE FOR THE GIVEN PARABOLA.
3)DERIVE THE EQUATION FOR REGULAR VELOCITY AT ANY INSTANT AND ALSO GIVE ITS DIRECTION.
Answers
step-by-step explanation:
1) The path of the projectile is shown in the figure.
The horizontal component of velocity always remain same as there is no acceleration on the horizontal axis.
Thus, the horizontal distance travelled by the projectile is
x = u cosθ × t
t = x/(u cosθ) ...... (1)
The vertical distance travelled by projectile is given from kinematical equation s = ut + ½at2.
Here, s = y, a = ay = -g, u = uy = u sinθ
Thus, we have
y = u sinθ -½ gt^2 ......... (2)
Substituting equation (1) in (2),
we get
y=xtanθ−gx^2/2v^2cos^2(θ)
Where,
y is the horizontal component,
x is the vertical component,
g= gravity value,
v= initial velocity,
θ = angle of inclination of the initial velocity from horizontal axis,
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2. (i) Time of Flight
In y -direction,
u = usinθ
a = −g
v = 0 (at highest point)
Now, Using the first equation of motion,
v = u + at
0 = usinθ − gt
⇒ gt = usinθ
⇒ t = usinθ/g
Since t is only half of the total time, therefore we can calculate total time of the journey as
T = 2t
⇒T= 2usinθ/g
(ii) Horizontal Range
Since, the projectile is in the air for a duration T
R = u in x direction × T
⇒R = ucosθ × 2usinθ/g
⇒R=u² 2sinθcosθ/g
∴ R = u²sin2θ/g
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3. (i) Component of initial velocity along x-axis.
=> ux = u1.
Component of initial velocity along y-axis.
=> uy = 0
(ii) Acceleration along x-axis
=> ax=0
(Because no force is acting along the horizontal direction)
Acceleration along y-axis
ay =g = 9.8m/s^2
It is directed downwards.
(iii) Component of velocity along the x-axis at any instant t.
=> vx=ux + axt = u + 0
=>vx=u
This means that the horizontal component of velocity does not change throughout the projectile motion.
Component of velocity along the y-axis at any instant t.
=>vy= uy + ayt = 0 + gt
=>vy=gt
(iv) The displacement along x-axis at any instant t
=> x=uxt + (1/2) axt^2
=> x=uxt + 0
=> x=u t
The displacement along y-axis at any instant t
y= uyt + (1/2) ayt^2
y= 0 + (1/2) ayt^2
y=1/2gt^2
Equation of a trajectory(path of a projectile)
We know at any instant x = ut
=> t=x/u
Also, y= (1/2)gt^2
Subsituting for t we get
y= (1/2)g(x/u)^2
=> y= (1/2)(g/u2)x^2
y= kx^2 where k= g/(2u^2 )
This is the equation of a parabola which is symmetric about the y-axis.Thus,the path of projectile,projected horizontally from a height above the ground is a parabola.
