Kinematics Ques 51
- The trajectory of a projectile in a vertical plane is $y=a x-b x^{2}$, where $a, b$ 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 ……… and the angle of projection from the horizontal is ……..
(1997C, 1M)
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Answer:
Correct Answer: 51.$a^{2} / 4 b, \tan ^{-1}(a)$
Solution:
Formula:
- $y=a x-b x^{2}$
For height (or $y$ ) to be maximum.
$$ \frac{d y}{d x}=0=a-2 b x \cdots(i) $$
$$ \therefore \quad x=\frac{a}{2 b} $$
(i) $\therefore y _{\max }=$ maximum height
$$ H=a(a / 2 b)-b(a / 2 b)^{2} $$
$$ H=a^{2} / 4 b $$
(ii) $\frac{d y}{d x} \underset{(x=0)}{ }=a=\tan \theta _0$
where, $\theta _0$ is the angle of projection.
$$ \therefore \quad \theta _0=\tan ^{-1}(a) $$
BETA
