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3-2.Motion in Plane
hard
A stone is projected from ground at $t = 0$. At the time of projection horizontal and vertical component of velocity are $10\, m/s$ and $20\, m/s$ respectively. Then time at which tangential and normal acceleration magnitude will be equal $(g = 10\, m/s^2)$ [neglect air friction] ......... $\sec$
A
$\frac{1}{2} $
B
$\frac{3}{2} $
C
$3$
D
$4$
Solution
at any instant
$a_{t}=g \sin \theta$
$a_{n}=g \cos \theta$
$\because a_{n}=a_{r}$
$\Rightarrow \tan \theta=1 \Rightarrow\left|\frac{v_{y}}{v_{x}}\right|=1$
$\Rightarrow \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\pm 1 \Rightarrow \frac{\mathrm{u}_{\mathrm{y}}-\mathrm{gt}}{\mathrm{u}_{\mathrm{x}}}=\pm 1$
Solving we get
$t=1$ and $3$
Standard 11
Physics
