Question

A particle is projected with velocity $20\sqrt{2}m/sat{45}^{\circ }$ with horizontal. After $1$ s find tangential and normal acceleration of the particle. Also, find radius of curvature of the trajectory at that point.

• A. $a_t=-2\sqrt{5}m{s}^{-2}$
• B. $a_n=4\sqrt{5}m{s}^{-2}$
• C. $R=25\sqrt{5}m$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a_n=4\sqrt{5}m{s}^{-2}$
B $R=25\sqrt{5}m$
C $a_t=-2\sqrt{5}m{s}^{-2}$

Given : $u=20\sqrt{2}$ $m/s$ $t=1$ s

$u_x=ucos{45}^o=20$ $m/s$

x direction : $a_x=0$ $⟹V_x=u_x=20$ m/s

Also $u_y=usin{45}^o=20$ m/s

y direction : $V_y=u_y-gt$

$\therefore$ $V_y=20-10\left(1\right)=10$ m/s

From figure $tan\alpha =\frac{V_y}{V_x}=\frac{10}{20}$

$\therefore tan\alpha =\frac{1}{2}$ $⟹sin\alpha =\frac{1}{\sqrt{5}}$ and $cos\alpha =\frac{2}{\sqrt{5}}$

Speed at that instant $V=\sqrt{V_x^2+V_y^2}=\sqrt{(20{)}^2+(10{)}^2}=\sqrt{500}=10\sqrt{5}$ m/s

Also from figure, radial acceleration $a_r=gcos\alpha =10\times \frac{2}{\sqrt{5}}=4\sqrt{5}$ $m/s^2$

Tangential acceleration $a_t=-gsin\alpha =-10\times \frac{1}{\sqrt{5}}=-2\sqrt{5}$ $m/s^2$

Let the radius of curvature at that instant be $R$

$\therefore$ $R=\frac{V^2}{a_r}=\frac{500}{4\sqrt{5}}=25\sqrt{5}$ m