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Question

In a projectile motion let vx and vy be the horizontal and vertical components of velocity at any time t and x and y be the displacements along horizontal and vertical from the point of projection at any time t. Then

A
vyt graph is a straight line with negative slope and positive intercept.
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B
xt graph is a straight line passing through origin.
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C
yt graph is a straight line passing through origin.
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D
vxt graph is a straight line parallel to t axis.
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Solution

The correct options are
A vyt graph is a straight line with negative slope and positive intercept.
B xt graph is a straight line passing through origin.
D vxt graph is a straight line parallel to t axis.
If u is the initial speed and θ is the angle of projection, then velocity of projectile along vertical direction is given by
vy=usinθgt
vyt graph is a straight line with negative slope and positive intercept.
Position of projectile at time t along horizontal direction is given by
x=(ucosθ)t
xt graph is a straight line passing through origin.
Position of projectile at any time t along the vertical direction is given by
y=(usinθ)t+12gt2
yt graph is a parabola.
Since projectile has uniform velocity along horizontal direction, we can say that
vx=ux=ucosθ
vxt graph should be a straight line parallel to t axis.
Thus, options (a), (b) and (d) are correct answers.

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