Question

The trajectory of a projectile in a vertical
plane is y= ax – Bx? where a and ß are
constants and x and y are respectively the
horizontal and vertical distances of the
projectile from the point of projection. The
maximum height attended by projectile is
(B)
2016
ole ola
(D)​

Answer 1

$\displaystyle\large\boxed {\sf {H=\dfrac {A^2}{4B}}}$

The trajectory is given as,

$\displaystyle\longrightarrow\sf {y=Ax-Bx^2}$

But it's actually,

$\displaystyle\longrightarrow\sf{y=x\tan\theta-\dfrac {g}{2u^2\cos^2\theta}x^2}$

Thus we get,

• $\displaystyle\sf {A=\tan\theta=\dfrac {\sin\theta}{\cos\theta}}$

• $\displaystyle\sf{B =\dfrac {g}{2u^2\cos^2\theta}}$

We know that the maximum height,

$\displaystyle\longrightarrow\sf{H=\dfrac {u^2\sin^2\theta}{2g}}$

$\displaystyle\longrightarrow\sf{H=\dfrac {\sin^2\theta}{\left (\dfrac {2g}{u^2}\right)}}$

$\displaystyle\longrightarrow\sf{H=\dfrac {\sin^2\theta}{\left (\dfrac {4g\cos^2\theta}{2u^2\cos^2\theta}\right)}}$

$\displaystyle\longrightarrow\sf{H=\dfrac {\sin^2\theta}{\left (4\cos^2\theta\cdot\dfrac {g}{2u^2\cos^2\theta}\right)}}$

$\displaystyle\longrightarrow\sf{H=\dfrac {\tan^2\theta}{\left (4\cdot\dfrac {g}{2u^2\cos^2\theta}\right)}}$

$\displaystyle\longrightarrow\sf{\underline{\underline {H=\dfrac {A^2}{4B}}}}$