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3-2.Motion in Plane
hard
For an object projected from ground with speed $u$ horizontal range is two times the maximum height attained by it. The horizontal range of object is ..........
A
$\frac{2 u^2}{3 g}$
B
$\frac{3 u^2}{4 g}$
C
$\frac{3 u^2}{2 g}$
D
$\frac{4 u^2}{5 g}$
Solution
(d)
$R=24$ also,$\frac{H}{R}=\frac{1}{4} \tan \theta$
$\frac{H}{R}=\frac{1}{2} \Rightarrow \frac{1}{2}=\frac{1}{4} \tan \theta$
$\tan \theta=2=\frac{P}{B}$
$R=\frac{2 u^2 \sin \theta \cos \theta}{g}$
$R=\frac{2 u^2}{g} \cdot \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}$
$R=\frac{4 u^2}{5 g}$
Standard 11
Physics
Similar Questions
Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following with $g=10\,m / s ^2$.
| $Column-I$ | $Column-II$ |
| $(A)$ Angle of projection | $(p)$ $20\,m$ |
| $(B)$ Angle of velocity with horizontal after $4\,s$ | $(q)$ $80\,m$ |
| $(C)$ Maximum height | $(r)$ $45^{\circ}$ |
| $(D)$ Horizontal range | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
medium
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