A projectile is given an initial velocity of - - 2- The cartesian equation of its path is g - 10 ms-2-

The correct option is B y = 2x 5x^2 tanθ = usinθu cosθ = 21 The desired equation is y = xtanθ gx^22u^2cos^2θ = x × 2 10x^22(√(2^2 + 1^2))^2 ...

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Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

A projectile ...

Question

A projectile is given an initial velocity of $^i + 2^j$ . The cartesian equation of its path is $(g = 10 m s^{- 2})$ .

y = 2 x - 5 x^{2}

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y = x - 5 x^{2}

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4 y = 2 x - 5 x^{2}

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y = 2 x - 25 x^{2}

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Solution

The correct option is B $y = 2 x - 5 x^{2}$
$tan θ = \frac{u sin θ}{u cos θ} = \frac{2}{1}$
The desired equation is
$y = x tan θ - \frac{g x^{2}}{2 u^{2} {cos}^{2} θ} = x \times 2 - \frac{10 x^{2}}{2 (\sqrt{2^{2} + 1^{2}})^{2} {(\frac{1}{\sqrt{5}})}^{2}}$
or $y = 2 x - 5 x^{2}$ .

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A projectile is given an initial velocity of ^i+2^j. The cartesian equation of its path is (g=10 ms−2).

The correct option is B y=2x−5x2tanθ=usinθucosθ=21The desired equation isy=xtanθ−gx22u2cos2θ=x×2−10x22(√22+12)2(1√5)2or y=2x−5x2.

A projectile is given an initial velocity of $^i + 2^j$ . The cartesian equation of its path is $(g = 10 m s^{- 2})$ .

The correct option is B $y = 2 x - 5 x^{2}$
$tan θ = \frac{u sin θ}{u cos θ} = \frac{2}{1}$
The desired equation is
$y = x tan θ - \frac{g x^{2}}{2 u^{2} {cos}^{2} θ} = x \times 2 - \frac{10 x^{2}}{2 (\sqrt{2^{2} + 1^{2}})^{2} {(\frac{1}{\sqrt{5}})}^{2}}$
or $y = 2 x - 5 x^{2}$ .