Question

.Explanation:
The trajectory of a projectile in a vertical plane is y=ax-bx2,where a and b are constants,and x and y are respectievly the horizontal and the verticaldistanceof the projectile from the point of projection.The maximum height attained is?and the angle of projectionfrom the horizontal is?

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Answer 1

$\huge\boxed{\blue {\rm solution }}$

given trajectory equation in vertical plane is:

y=ax-b^2

let us differentiate it with respect to x.

$\tt \dfrac{dy}{dx}=\dfrac{d (ax-bx^2)}{dx}$

$\tt =a-2bx$

we'll use the concept of maxima and minima to find the minimum and maximum heights of the projection.

$\tt a-2bx =0$

$\tt x=\dfrac {a }{2b }$

the maximum height attained will be:

$\tt y=ax-b^2$

$\tt y= a^2/4b$

we know that the equation of trajectory is given by the formula:

$\boxed{\red {\tt xtanA-\dfrac {g}{2u^2cos^2A}×x^2 }}$

comparing our trajectory equation to thi equation, the coefficient of x is a and tani respectively. from this we can conclude that tanA=a

so the angle made by the projection from the horizontal will be tan inverse of angle(a)

hope my solution helps you out!

Answer 2

Answer:

hilu

Step-by-step explanation:

The trajectory of a projectile in a vertical plane is y=ax−bx2, where a, b are constants, and x and y are, respectively, the horizontal and vertical distance of the projectile from the point of projection.

mark me dr ❤️