Saved Bookmarks
| 1. |
A projectile is fired horizontally with a velocity u . Show that its trajectory is a parabola. Also obtain the expression for. Horizontal range . |
|
Answer» Solution :Let us consider a projectile, that is, thrown horizontalwith an initialvelocity `vec(u)`from the topof a tower of height h in the figure . MOTION along horizontal DIRECTION :The particle has zero acceleration along x direction. So, the initial velocity `u(x)`REMAINS constant throughout the motion. The distance traveled by the projectile at a time t is GIVENBY `x = u_(x) t + (1)/(2) at^(2)` Since a = 0 along x direction, we get `x =u_(x)t "". . . (1)` Motion along downward direction : Here `u_(y) = 0 ` (initial velocity has no downward component), a = g .The distance y at time y is given by `y = u_(y) t + (1)/(2) at ^(2)`, we get `y = (1)/(2) g t^(2) `[u = 0, a = g]. . . (2) Substituting the value of t from equation (1) in equation (2) we have `y = (1)/(2) g""(x^(2))/(u_(x)^(2))= ((g)/(2u_(x)^(2)))x^(2)` `y = Kx^(2) "" . . . (3)` Where `K = (g)/(2u_(x)^(2))` is a constant of proportionality . Equation (3) GIVES the trajectory of the projectile. Thus , the path followed by the projectile is a parabola . Horizontal range :The horizontal distance coveredby the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range.For horizontal motion, we have `s_(x) = u_(x) t + (1)/(2) at^(2)` Here, `s_(x) = R`(range),`u_(x) = u, a = 0 ` (no horizontal acceleration) T is time of flight . Then horizontal range = u T. Since the time of flight T ` = sqrt((2h)/(g))`, we substitute this and we get the horizontal range of the particle as`R = u sqrt((2h)/(g))`. 
|
|