A projectile is given an initial velocity of - - 2-m-s- where - is along ground and - is along the vertical- If g - 10 m-s2- the equation of its trajectory is

The correct option is D y =2x 5x^2 The equation of projectile is given by, Y=xtanθ gx^22u^2cos^2θ tanθ=u_yu_x=2/1=2 ⇒ y=2x 1/210x^2(1^2+2^ ...

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Standard XII

Physics

The Equation for the Path of Projectile

A projectile ...

Question

A projectile is given an initial velocity of $(^i + 2^j) m / s$ , where $^i$ is along ground and $^j$ is along the vertical. If $g = 10 m / s^{2}$ , the equation of its trajectory is

y = 2 x - 5 x^{2}

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

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4 y = 2 x - 25 x^{2}

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y = x - 5 x^{2}

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Solution

The correct option is D $y = 2 x - 5 x^{2}$
The equation of projectile is given by,
$Y = x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}$
$t a n θ = {\frac{u_{y}}{u}}_{x} = \frac{2}{1} = 2$
$\Rightarrow y = 2 x - \frac{1}{2} \frac{10 x^{2}}{(1^{2} + 2^{2}) \times \frac{1}{5}}$
$\Rightarrow y = 2 x - 5 x^{2}$
Hence the correct option is A

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A projectile is given an initial velocity of (^i+2^j)m/s, where ^i is along ground and ^j is along the vertical. If g=10 m/s2, the equation of its trajectory is

The correct option is D y=2x−5x2The equation of projectile is given by,Y=xtanθ−gx22u2cos2θtanθ=uyux=21=2⇒y=2x−1210x2(12+22)×15⇒y=2x−5x2Hence the correct option is A

A projectile is given an initial velocity of $(^i + 2^j) m / s$ , where $^i$ is along ground and $^j$ is along the vertical. If $g = 10 m / s^{2}$ , the equation of its trajectory is