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Class 11>>Physics>>Motion in a Plane>>Projectile Motion>>Prove that the path of projectile motion
Question
Prove that the path of projectile motion in parabolic.
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Solution
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path of a projectile
let OX be a horizontal line on the ground and OY be a verticle line O is the origin for X and Y axis Consider that the projectile is fired with velocity u and making angle θ with the horizontal form the point ′O′ on the ground
The velocity of projection of the projectile can be resolved into the following two components
(i) ux=ucosθ,alongOX
(ii) uy=usinθ,alongOY
As the projectile moves it covers distance along the horizontal due to the horizontal component ucosθ of the velocity of projection and along verticle due to the vertical component usinθ . let that any time t , the projectile reaches the point P , So that its distance along the X and Y axis are given by x and y respetively.
Motion along horizontal direction : it we neglect friction due to air, then horizontal component of the velocity i.e. ucosθ will remain constant. thus
Initial velocity along the horizontal ux=ucosθ
Acceleration along the horizontal ax=0
The position of the projectile along X-axis at any time t is given by
x=uxt+21axt2
Putting ux=ucosθ and ax=0 we have
x=(ucosθ)t+21(o)t2
or x=(ucosθ)t
or t=
Explanation:
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Answer:
The path followed by a projectile is called its trajectory.
Explanation:
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