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Question

The trajectory of a projectile in a vertical plane is y=ax−bx2, where a and b are constants and x and y are horizontal and vertical distances respectively of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are:

A

b22a,tan1(a)

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B

a2b,tan1(2a)

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C

a24b,tan1(a)

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D
2a2b,tan1(a)
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Solution

The correct option is C

a24b,tan1(a)


Given,
y=axbx2

Comparing with

y=(tanθ)xg(1+tan2θ)x22u2

we get:

(i) a=tanθ


sinθ=a1+a2

Also, θ=tan1a

(ii) b=g(1+tan2θ)2u2

Substituting the value, tanθ=a, we have,

u2=g(1+a2)2b

Now, the maximum height of the given projectile,

hmax =u2sin2θ2g

Substituting the values in the above equation, we get

hmax=(1+a2)4ba21+a2

hmax=a24b

Hence, option (c) is correct answer.


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