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Standard XII

Physics

Magnitude of Instantaneous Velocity

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 = (8t - 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 - 10t, $v_{x}$ = $\frac{d x}{d t}$ = 6
at the time of projection i.e. $v_{y}$ = $\frac{d y}{d t}$ =8 and $v_{x}$ = 6
$∴$ v = $\sqrt{{(v_{x})}^{2} + {(v_{y})}^{2}}$ = $\sqrt{{(6)}^{2} + {(8)}^{2}}$ = 10 m/s

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Similar questions

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

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

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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 = √(vx)2+(vy)2 = √(6)2+(8)2 = 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 = (8t - 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 - 10t, $v_{x}$ = $\frac{d x}{d t}$ = 6
at the time of projection i.e. $v_{y}$ = $\frac{d y}{d t}$ =8 and $v_{x}$ = 6
$∴$ v = $\sqrt{{(v_{x})}^{2} + {(v_{y})}^{2}}$ = $\sqrt{{(6)}^{2} + {(8)}^{2}}$ = 10 m/s