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Trajectory of particle in a projectile m...

Trajectory of particle in a projectile motion is given as `y = x - (x^(2))/(80)`. Here x and y are in metre. For this projectile motion match the following with `g = 10 ms^(-2)`.

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The correct Answer is:
`A rarr r,B rarr r,C rarr p,D rarr q`
Comparing with the standard equation of projectile,
`y = x tan theta - (gx^(2))/(2u^(2)cos theta)`
We get, `theta = 45^(@)` and `u = 20 sqrt(2)ms^(-1)`
Time period of this projectile is 4s. Hence, after 4s, velocity vector will again make and angle of `45^(@)` with horizontal.
So, maximum height, `h = (u^(2)sin^(2)theta)/(2g) = ((20sqrt(2))^(2)sin^(2)(45^(@)))/(2 xx 10)`
`H = (400 xx 2 xx (1)/(2))/(20) = (400)/(20) = 20m`
and horizontal range, `R = (u^(2)sin 2 theta)/(g) = ((20sqrt(2))^(2)sin2(45^(@)))/(10)`
`=(400 xx 2 xx1)/(10) ( :' sin 90^(@) = 1)`
`rArr R = 80 m`
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