Question

A stone is projected from ground as shown in figure at t=0. Then find minimum time (in sec) after which magnitude of tangential and normal acceleration become same?
uy= 20 m/s
ux= 10 m\s
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Answer 1

The time is 3 seconds

Answer 2

Given info : A stone is projected from ground as shown in figure at t = 0. where horizontal and vertical components of velocity are 10 and 20 m/s respectively.

To find : the minimum time (in sec) after which the magnitude of tangential and normal acceleration become same is ..

solution : for projectile motion,

tangential acceleration, $a_t$ = gsinθ

normal acceleration, $a_n$ = gcosθ

if we assume θ is an angle made by stone at that time when tangential and normal acceleration are same.

here gsinθ = gcosθ

⇒tanθ = 1

⇒θ = 45°

now, at that time we can apply, tanθ = $\left|\frac{v_y}{v_x}\right|$

⇒1 = $\left|\frac{v_y}{v_x}\right|$

⇒$|v_x|=|v_y|$

we know, horizontal component of velocity remains constant. so $v_x=u_x=10m/s$

vertical component at time t, $v_y=u_y-gt$

= 20 - 10t

now, |10| = |20 - 10t|

⇒±10 = 20 - 10t

⇒t = 1, 3

therefore minimum time is 1s after which tangential and normal acceleration are same.