So at point P, v x = u ……………… (i)Ĭonsidering motion along vertical axis, v y = u y + g t As, initial velocity along Y axis = u y = 0, hence, v y = g t …………(ii) Magnitude of resultant velocity at any point P v 2 = v x 2 + v y 2 => So the equation of the horizontal projectile velocity is: v = √(v x 2 + v y 2) = √(u 2 + g 2 t 2) ……………. The velocity equation of the horizontal projectile | derive the equation of horizontal projectile velocityĪlong the horizontal axis,a x =0 so, velocity remains constant along horizontal axis. So, the trajectory of the projectile launched parallel to the horizontal is a parabola. (1) Now, Considering motion along Y axis, u y = 0 (initial velocity along Y axis, at t = 0) a y = g Now, distance traveled along the Y-axis can be expressed as: y = u y t + (½) g t 2 => y = (½) g t 2 ……………… (2) From 1 & 2 we get, y = (½) g t 2 => y = (½) g (x/u) 2 => y = x 2 Say, after a time duration of t the projectile reaches point P (x,y).Ĭonsidering motion with uniform velocity along X-axis, distance traveled along the X-axis in time t can be expressed as: Trajectory equation of the horizontal projectile & its derivation So in the equations, we will refer to the downward acceleration ay of the projectile as +g. In this figure Y-axis is taken downwards, therefore, the downward direction will be regarded as the positive direction. (ii) vertically downward accelerated motion with constant acceleration g (due to gravity). Now the body moves along path OPA under the influence of two simultaneously independent motions listed below: (i) motion with uniform horizontal velocity u. h is the height of Point O with respect to the ground. Suppose a body is thrown horizontally from a point O with velocity u. Figure 1: Horizontal projectile and derivation of its equations
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