Question

Show that the path of projectile projected at an angle of THITA from horrizontal is parabolic in shape?

Answer 1

If the projectile is projected with a velocity `u` at an angle `theta` w.r.t horizontal,then horizontal component of its velocity is `u cos theta` due to which it moves forward,and vertical component of velocity is `u sin theta` due to which first it goes up then comes down due to gravity.

So,if in time `t` it goes `X` distance horizontally ,we can say,

`(X = u cos theta*t)`

Or, `t=X/(u cos theta)`...1

And,if vertically it goes by distance `Y` then,

`Y=u sin theta×t - 1/2 g t^2`

Now putting the value of `t` in the above equation we get,

`Y= X tan theta - 1/2 g X^2/(u^2 cos ^2 theta)`

This can be compared with `y=ax+bx^2` ,which is an equation of parabola