An object is projected with a velocity of $20 \mathrm{~m} / \mathrm{s}$ making an angle of $45^{\circ}$ with…

An object is projected with a velocity of $20 \mathrm{~m} / \mathrm{s}$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory is $h=A x-B x^2$ where $h$ is height, $x$ is horizontal distance $A$ and $B$ are constants. The ratio $A: B$ is $\left(g=10 \mathrm{~ms}^{-2}\right)$
  1. $1: 5$
  2. $5: 1$
  3. $1: 40$
  4. $40: 1$

Solution

$u=20 \mathrm{~m} / \mathrm{s}, \theta=45^{\circ}$ Equation of trajectory $h=A x-B x^2$ Standard equation $h=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$ On comparing the two equations $\begin{aligned} A & =\tan \theta, B=\frac{g}{2 u^2 \cos ^2 \theta} \\ \frac{A}{B} & =\frac{\tan \theta}{\frac{g}{2} u^2 \cos ^2 \theta}=\frac{2 u^2 \cos ^2 \theta \tan \theta}{g} \\ & =\frac{2 \times(20)^2 \cos ^2 45^{\circ} \tan 45^{\circ}}{10} \\ & =\frac{2 \times 400 \times\left(\frac{1}{\sqrt{2}}\right)^2 \times 1}{10} \\ & =2 \times 40 \times \frac{1}{2}=40\end{aligned}$ $A: B=40: 1$

Asked in: AP EAMCET 2001

Practice more Motion In Two Dimensions questions on Aicharya