A particle is projected at angle `theta` with horizontal with velocity `v_0 at t = 0`. Find
(a) tangential and normal acceleration of the particle at `t = 0` and at highest point of its trajectory.
(b) the radius of curvature `att = 0` and highest point.
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Explanation

(a) The direction of tangential acceleration is in the line of velocity and the direction perpendicular to velocity direction. The tangential and normal directions at `O and P` are shown in (Figs. 161 and 5.162), respectively.
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 = -g sin theta`
Normal acceleration,
`a_n = g cos theta`
At P(at highest point) :
Tangential acceleration,
`a_t = 0 and a_n = g`
(b) Let the radius of curvature at `O` be `R_0`. The normal acceleration at `O be g cos theta`.
`a_n = (v^2)/(R) rArr R_0 = (v_0^2)/(a_n) = (v_0^2)/(g cos theta)`
Radius of curvature at `P` :
Normal acceleration at `P` is `g`. Hence,
`R_P = (v^2)/(a_n) = ((v_0 cos theta)^2)/(g) = (v_0^2 cos^2 theta)/(g)`.
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