Physics · Motion In A Plane · NEET
After we eliminate time t, we get y = (tan theta) x - [g/(2 u^2 cos^2 theta)] x^2. Here tan theta, u and g are fixed numbers, so the equation looks like y = ax - bx^2 (a constant times x, minus a constant times x squared). Any equation where y depends on x and x^2 (with no higher power) is, by definition, a parabola. So a projectile MUST trace a parabola. It is not a circle, because a circle needs x^2 and y^2 together; here only x is squared.
In projectile motion we first write two separate equations: x = (u cos theta) t and y = (u sin theta) t - (1/2) g t^2. These both contain time t. To find the SHAPE of the path (y in terms of x), we do not want t in the answer. So we solve the x equation for t, getting t = x / (u cos theta), and substitute it into the y equation. This removes t and leaves a direct relation between y and x. That is 'eliminating time.'
Start: x = (u cos theta) t, so t = x/(u cos theta). Put this into y = (u sin theta) t - (1/2) g t^2. First term: (u sin theta)(x/(u cos theta)) = (tan theta) x. Second term: (1/2) g (x/(u cos theta))^2 = g x^2 / (2 u^2 cos^2 theta). So y = (tan theta) x - g x^2 / (2 u^2 cos^2 theta). Done.
Yes, it is simpler. For a horizontal throw the launch angle theta = 0, so the vertical launch speed is zero and there is no (tan theta) x term. Taking downward as positive, y = g x^2 / (2 u^2). This is still an x^2 relation, so it is still half of a parabola. The full formula y = (tan theta)x - gx^2/(2u^2cos^2theta) reduces to this when theta = 0.
The first term (tan theta) x is the straight line the projectile WOULD follow if there were no gravity, going up along the launch direction. The second term, minus g x^2 / (2 u^2 cos^2 theta), is how far gravity pulls the body BELOW that straight line. This 'drop' grows with x^2, so the path bends down more and more, forming the parabola.
The x and y coordinates of a particle at any time t are x = 5t - 2t^2 and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
y = (tan theta) x - g x^2 / (2 u^2 cos^2 theta), where u is the launch speed, theta is the angle of projection with the horizontal, and g is the acceleration due to gravity.
After eliminating time, y depends on x and x^2 only, with all other quantities constant. An equation of the form y = ax - bx^2 is the standard equation of a parabola, so the path is a parabola.
Yes. The parabolic result assumes no air resistance and constant gravity. With air resistance the real path is not a perfect parabola, but for NEET we ignore air resistance unless told otherwise.
For theta = 0, it becomes y = g x^2 / (2 u^2), taking downward as positive. This is still an x^2 relation, so the path is half a parabola.
For a given projectile, u, theta and g are constants. Only x and y vary. That is why the equation reduces to the parabola form y = ax - bx^2.
