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Standard XII

Physics

Applications of Horizontal and Vertical Components

The horizonta...

Question

The horizontal distance x and the vertical height y of a projectile at a time t are given by
$x = a t a n d y = b t^{2} + c t$
where a, b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?

$\sqrt{a^{2} + (2 b + c)^{2}}$

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$\sqrt{2 a^{2} + (b + c)^{2}}$

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$\sqrt{2 a^{2} + (2 b + c)^{2}}$

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$\sqrt{a^{2} + (b + 2 c)^{2}}$

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Solution

The correct option is A
$\sqrt{a^{2} + (2 b + c)^{2}}$

The horizontal component of velocity is
$v_{x} = \frac{d x}{d t} = \frac{d}{d t} (a t) = a . . . . . . . . (1)$
The vertical component of velocity is
$v_{y} = \frac{d y}{d t} = \frac{d}{d t} (b t^{2} + c t) = 2 b t + c . . . . . . . . (2)$
The value of $v_{y}$ at t = 1 s is (2b + c). Therefore, the magnitude of velocity at t = 1 s is
$v = \sqrt{(v_{x}^{2} + v_{y}^{2})} = \sqrt{a^{2} + (2 b + c)^{2}}$
Thus, the correct choice is (a).

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The horizontal distance x and the vertical height y of a projectile at a time t are given by x=at and y=bt2+ct where a, b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?

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$x = a t a n d y = b t^{2} + c t$
where a, b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?