A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacement x and y (in metres) respectively vary with time t
(in seconds) as
What is the magnitude and direction of the velocity with which the body is projected?
20 ms –1 at an angle of 30o with the horizontal
We know that the position coordinates x and y are given by
Comparing Eq. (i) with we have,
Also, comparing Eq. (ii) with y = 10t – t2, we have, v0 sin θ = 10 ms–1. These equations give which gives θ = 30o.
It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point P at a distant r from the point of projection. The product of the times taken to reach this point in the two possible ways is then proportional to
At what angle (θ) with the horizontal should a body be projected so that its horizontal range equals the maximum height it attains?
A body is projected horizontally from a point above the ground. The motion of the body is described by the equations
x = 2t
and y = 5 t2
Where x and y are the horizontal and vertical displacements (in m) respectively at time t. The trajectory of the body is