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Standard XII

Physics

The Equation for the Path of Projectile

The height y ...

Question

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8 t - 5 t^{2})$ meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is

8 m/sec

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6 m/sec

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10 m/sec

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Solution

The correct option is C 10 m/sec
$v_{y} = \frac{d y}{d t} = 8 - 10 t, v_{x} = \frac{d x}{d t} = 6$ at the time of projection i.e. $v_{y} = \frac{d y}{d t} = 8 a n d v_{x} = 6$
$∴ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$

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Q. The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by

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The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y=(8t−5t2) meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is

The correct option is C 10 m/secvy=dydt=8−10t, vx=dxdt=6 at the time of projection i.e. vy=dydt=8 and vx=6 ∴ v=√v2x+v2y=√62+82=10 m/s

The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8 t - 5 t^{2})$ meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is

The correct option is C 10 m/sec
$v_{y} = \frac{d y}{d t} = 8 - 10 t, v_{x} = \frac{d x}{d t} = 6$ at the time of projection i.e. $v_{y} = \frac{d y}{d t} = 8 a n d v_{x} = 6$
$∴ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$