![]() ![]() A function assigns exactly one output to each input of a specified type. In our example above, x is the independent variable and y is the dependent variable.Ī function is an equation that has only one answer for y for every x. The variable which we assign the value we call the independent variable, and the other variable is the dependent variable, since it value depends on the independent variable. In our equation y=x+7, we have two variables, x and y. We could instead have assigned a value for y and solved the equation to find the matching value of x. If we would have assigned a different value for x, the equation would have given us another value for y. Velocity with which the projectile hits the ground :Īt A, v = √(v x 2 + v y 2) At A, v x=u….If we in the following equation y=x+7 assigns a value to x, the equation will give us a value for y. So, R=Horizontal velocity×Time of flight= u×T=u√(2h/g) Hence, Range of a horizontal projectile = R = u√(2h/g) ![]() Here we will use the equation for the time of flight, i.e. It is the horizontal distance covered by the projectile during the time of flight. => h = 0 + (½) g T 2 Time of flight of a horizontal projectile = T = √(2h/g) …………… (4) Range of a horizontal projectile For the vertical downward movement of the body we use this equation: s y = u y + (½) g t 2 It is the total time for which the projectile remains in flight (from O to A). Now if β is the angle which resultant velocity makes with the horizontal, then tanβ = v y/v x= gt / u => β = tan -1(gt/u) Time of flight of a projectile launched horizontally 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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