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Question

The trajectory of a projectile in a vertical plane is: y=axbx2,
where a and b are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is

A
a24b
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B
a28b
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C
a22b
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D
a22gb
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Solution

The correct option is A a24b
The equation of trajectory is, y=axbx2
For maximum height of projectile, y should be maximum.
dydx=0 and d2ydx2<0
dydx=a2bx=0
x=a2b
d2ydx2=2b<0
For x=a2b, y is maximum
Substituting the value of x=a2b in trajectory equation,
ymax=a(a2b)b(a2b)2
ymax=a24b

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