Question

4. a body is projected into air at an angle θ with the horizontal. a) what is the trajectory followed by this projectile? i) ellipse ii) parabola iii) straight line iv) circle (1) b) give a mathematical proof for your answer. (2) c) trajectory of a body in a projectile motion is given by 2 . 80 x y x   x and y are in meters. find maximum height of this projectile.

Answer 1

a body is projected with speed u at an angle $\theta$ with horizontal.

after time t,

velocity of body is given by,

$\vec{v}=ucos\theta\hat{i}+(usin\theta-gt)\hat{j}$

initial velocity of body is given by, $\vec{u}=ucos\theta\hat{i}+usin\theta\hat{j}$

distance covered by body in x direction,

$x=u_xt+\frac{1}{2}a_xt^2$

we know, acceleration in horizontal direction of projectile motion equals zero.

so, $a_x$ = 0, $u_x=ucos\theta$

so, $x=ucos\theta t$.......(1)

distance covered by body in y direction,

$y=u_yt+\frac{1}{2}a_yt^2$

here,$u_y=usin\theta$, $a_y=-g$

so, $y=usin\theta t-\frac{1}{2}gt^2$........(2)

from equations (1) and (2),

$y=usin\theta\frac{x}{ucos\theta}-\frac{1}{2}g\frac{x^2}{u^2cos^2\theta}$

$\boxed{\bf{y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}}}$

This expression shows trajectory followed by this particle is parabolic.

Answer 2

Answer:

The trajectory followed by this projectile is Parabolic.

Explanation:

PARABOLIC