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  • The trajectory of the projectile in the vertical plane is <img alt="y=ax-bx^2" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3Dax-bx%

The trajectory of the projectile in the vertical plane is y=ax-bx^2, where a and b are constant and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection.   The projectile angle on the horizontal plane is:

Option: 1

\theta=\tan^{-1}a


Option: 2

\theta=\cos^{-1}a


Option: 3

\theta=\sin^{-1}a


Option: 4

\theta=sec^{-1}a


Answers (1)

best_answer

Y=ax-bx^2

Comparing the given equation with:

Or, y=\left(tan\theta\right)x-\left(\frac{g}{2u^2\cos^2\theta}\right)x^2

Then, a=tan\theta

and, b=\frac{g}{2u^2\cos^2\theta}

Maximum height =   \frac{a^2}{4b} and angle of projectile:

Or, tan\theta=a

Or, \theta=\tan^{-1}a

Posted by

Rishi

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