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A heavy particle is projected with a vel...

A heavy particle is projected with a velocity at an angle with the horizontal into the uniform gravitational field. The slope of the trajectory of the particle varies as

A

.

B

.

C

.

D

.

Text Solution

Verified by Experts

The correct Answer is:
B, C
(b.,c.) Initially, accelerations are opposite to velocities. Hence, motion will be retarded. But after sometimes velocity will become zero and then velocity will in the direction of acceleration. Now the motion will be acceleration. As the particle is blown over by a wind with constant velocity along horizontal direction, the particle has a horizontal component of velocity. Let this component be `v_0`. Then it may be assumed that the particle is projected horizontally from the top of the tower with velocity `v_0`
Hence, for the particle, initial velocity `u = v_0` and angle of projection `theta = 0^@`.
We know equation of trajectory is
`y = x tan theta - (gx^2)/(2 u^2 cos^2 theta)`
Here, `y = -(gx^2)/(2 v_0^2)` `("putting" theta = 0^@)`
The slope of the trajectory of the particle is
`(dy)/(dx) = -(2gx)/(2 v_0^2) = -(g)/(v_0^2) x`
Hence, the curve between slope and `x` will be a straight line passing through the origin and will have a negative slope. It means that option (b) is correct.
Since horizontal velocity of the particle remains constant, `x = v_0 t`.
We get `(dy)/(dx) = - ("gt")/(v_0)`
So the graph between `m` and time `t` will have the same shape as the graph between `m and x`. Hence, option (a) is wrong.
The vertical component of velocity od the particle at time `t` is equal to `"gt"`. Hence, at time `t`,
`KE = (1)/(2)m [("gt")^2 + (v_0)^2]`
It means, the graph between `KE` and time `t` should be a parabola having value `(1)/(2) mv_0^2 at t = 0`. Therefore, option ( c) uis correct.
As the particle falls, its height decreases and `kE` increases.
`KE = (1)/(2) mv_0^2 + mg (H - h)`, where `H` is the initial height.
The `KE` increases linearly with height of its fall or the graph between `KE` and height of the particle will be a straight line having negative slope. Hence, option (d) is wrong.
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