A particle is thrown with initial velocity 8 m - s making angle 120- with constant acceleration vector 2 m - s 2 as shown- The time after speed again becomes 8 m - s isA- 4 sB- 8 sC- 2 sD- 1 s

The correct option is A 4s v_x = u_x + a_xt = 8 (2 sin 30^∘)t = 8 t v_y = u_y + a_yt = 0 + (2 cos 30^∘)t =√(3) t ∴ Velocity after a time 't' is v = √( ...

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Standard XII

Physics

Motion Under Variable Acceleration

A particle is...

Question

A particle is thrown with initial velocity 8 m/s making angle $120^{\circ}$ with constant acceleration vector $2 m / s^{2}$ as shown. The time after speed again becomes 8m/s is

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Solution

The correct option is C 4s

$v_{x} = u_{x} + a_{x} t = 8 - (2 s i n 30^{\circ}) t = 8 - t$

$v_{y} = u_{y} + a_{y} t = 0 + (2 c o s 30^{\circ}) t = \sqrt{3} t$

$∴$ Velocity after a time 't' is

$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

$v = \sqrt{(8 - t)^{2} + (\sqrt{3} t)^{2}}$

$8^{2} = (8 - t)^{2} + 3 t^{2} \Rightarrow t = 4 s$

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A particle is thrown with initial velocity 8 m/s making angle 120∘ with constant acceleration vector 2m/s2 as shown. The time after speed again becomes 8m/s is

The correct option is C 4s vx=ux+axt=8−(2sin30∘)t=8−t vy=uy+ayt=0+(2cos30∘)t=√3t ∴ Velocity after a time 't' is v=√v2x+v2y v=√(8−t)2+(√3t)2 82=(8−t)2+3t2 ⇒t=4s

A particle is thrown with initial velocity 8 m/s making angle $120^{\circ}$ with constant acceleration vector $2 m / s^{2}$ as shown. The time after speed again becomes 8m/s is