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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 vy−t graph is a straight line with negative slope and positive intercept. B x−t graph is a straight line passing through origin. D vx−t 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=usinθ−gt ∴ vy−t 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 ∴ x−t 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 ∴ y−t graph is a parabola. Since projectile has uniform velocity along horizontal direction, we can say that vx=ux=ucosθ ∴ vx−t graph should be a straight line parallel to t− axis. Thus, options (a), (b) and (d) are correct answers.

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