Question

Prove the that the projecticle motion palabolic​

Answer 1

Answer:

Let OX be a horizontal line on the ground and OY be a vertical line: O is the origin for X and Y axis. Consider that a projectile is fired with velocity u and making an angle θ with the horizontal from the point ‘O’ on the ground figure The velocity of projection of the projectile can be resolved into the following two components (i) ux = u cosθ, along OX (ii) uy = u sinθ, along OY. As the- projectile moves, it covers distance along the horizontal due to the horizontal component u cosθ of the velocity of projection and along vertical due to the vertical component u sinθ. Let that any time t, the projectile reaches the point P, so that its distances along the X and Y-axis are given by x and y respectively. Motion along horizontal direction: we neglect the friction due to air, then horizontal component of the velocity i. e., u cosθ will remain constant. Thus Initial velocity along the horizontal, ux = u cosθ Acceleration along the horizontal, ax = 0 The position of the projectile along X-axis at any time t is given by x = uxt + 1 2 12 axt2 Putting ux= u cosθ and ax = 0, we have x = (u cosθ)t + 1 2 12 (0)t2 or x = (ucosθ)t or t = x u cos θ xucos⁡θ ……………… (i) Motion along vertical direction : The velocity of the projectile along the vertical goes on decreasing due to effect of gravity Initial velocity along vertical, uy = u sinθ Acceleration along vertical, ay = -g The position of the projectile along T-axis at any time t is given by This is an equation of a parabola. Hence the path of a projectile projected at some angle with the horizontal direction is a parabolaRead more on Sarthaks.com - https://www.sarthaks.com/743890/prove-that-the-path-of-projectile-motion-is-parabolic