Question

A particle is projected at angle θ with horizontal with velocity $V_0$

Find

(i) tangential and normal acceleration of the particle at t = 0 and at highest point of its trajectory.

(ii) radius of curvature at t = 0 and highest point.

• A. at $t=0,a_t=-gsin\theta$
• B. At highest point $a_n=-g$
• C. at $t=0,R=\frac{V^2}{sin\theta }$
• D. At highest point $R=\frac{V^{2}co{s}^2\theta }{g}$

Answer 1

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

At highest point $R=\frac{V^{2}co{s}^2\theta }{g}$

(i) The direction of tangential acceleration is in the line of velocity and the normal acceleration is in the direction perpendicular to velocity direction. The tangential and normal directions at O & P are shown in figure.
The net acceleration of the particle during motion is acceleration due to gravity i.e., 'g' is acting vertically downward.


At O:(at t=0)
Tangential acceleration $a_t$=-$gsin\theta$
Normal acceleration $a_n$= $gcos\theta$
At P (at highest point)
Tangential acceleration
$a_t=0a_n=g$
(ii) Let radius of curvature at O be $R_0$. The normal acceleration at O is g cos $\theta$.
$a_n=\frac{v^2}{R}\Rightarrow R_0=\frac{v_0^2}{a_n}=\frac{v_0^2}{gcos\theta }$

Radius of curvature at P
Normal acceleration at P is 'g'
$R_p=\frac{v^2}{a_n}=\frac{(v_{0}cos\theta {)}^2}{g}=\frac{v_0^{2}co{s}^2\theta }{g}$