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
• B. At highest point
• C. at
• D. At highest point

Answer 1

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

The correct options are
A

at

B

At highest point

C

at

D

At highest point

(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'
$R_p=\frac{v^2}{a_n}=\frac{(v_{0}cos\theta {)}^2}{g}=\frac{v_0^{2}co{s}^2\theta }{g}$