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Motion in A Plane question

2022 · 26 Jul · Shift 1 · Q64
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  5. /2022 · 26 Jul · Shift 1 · Q64

Motion in A Plane question

2022 · 26 Jul · Shift 1 · Q64

JEE MainPhysicsMotion in A PlaneNumerical+4 / −1
If the initial velocity in horizontal direction of a projectile is unit vector i^\hat{i}i^ and the equation of trajectory is y=5x(1−x)y=5 x(1-x)y=5x(1−x). The yyy component vector of the initial velocity is ‾\underline{\hspace{2cm}}​j^\hat{j}j^​. (Take\mathrm{Take}Take g=10 m/s2)\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)g=10 m/s2)
Numerical answer
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Correct answer: 5

  1. Write the projectile trajectory equation

For a projectile launched with initial velocity components: uxi^+uyj^u_x \hat{i} + u_y \hat{j}ux​i^+uy​j^​ its trajectory is y=uyuxx−g2ux2x2.y = \frac{u_y}{u_x}x - \frac{g}{2u_x^2}x^2.y=ux​uy​​x−2ux2​g​x2.

  1. Given horizontal component

The initial velocity in horizontal direction is unit vector i^\hat{i}i^, so ux=1 m/s.u_x = 1\ \text{m/s}.ux​=1 m/s.

Also, g=10 m/s2g=10\ \text{m/s}^2g=10 m/s2.

So the trajectory becomes y=uyx−102⋅12x2=uyx−5x2.y = u_y x - \frac{10}{2\cdot 1^2}x^2 = u_y x - 5x^2.y=uy​x−2⋅1210​x2=uy​x−5x2.

  1. Compare with the given trajectory

Given: y=5x(1−x)=5x−5x2.y = 5x(1-x) = 5x - 5x^2.y=5x(1−x)=5x−5x2.

Comparing with y=uyx−5x2,y = u_y x - 5x^2,y=uy​x−5x2, we get uy=5.u_y = 5.uy​=5.

  1. Hence the yyy-component vector

The yyy-component of initial velocity is 5j^.5\hat{j}.5j^​.

So the required blank is 5.5.5.

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