A ball is thrown from the ground with a velocity of 20-3 m-s making an angle of 60-with the horizontal- The ball will be at a height of 40 m from the ground after a time t equal to g-10 ms-2

The correct option is C 2 sec As s=u sinθ t 12gt^2 40 = 20√(3)× (√(3)/2)t 12× 10× t^2 5t^2 30t+40=0 t=2 or 4 the minimum time t=2 ...

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Standard XII

Physics

Acceleration

A ball is thr...

Question

A ball is thrown from the ground with a velocity of $20 \sqrt{3}$ m/s making an angle of $60^{\circ}$ with the horizontal. The ball will be at a height of $40 m$ from the ground after a time $t$ equal to ( $g = 10 m s^{- 2}$ )

\sqrt{2} s e c

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\sqrt{3} s e c

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2 s e c

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3 s e c

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Solution

The correct option is C $2 s e c$
As $s = u sin θ t - \frac{1}{2} g t^{2}$
$40 = 20 \sqrt{3} \times (\sqrt{3} / 2) t - \frac{1}{2} \times 10 \times t^{2}$
$⟹ 5 t^{2} - 30 t + 40 = 0$
$⟹ t = 2 o r 4$
the minimum time $t = 2 s$

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A ball is thrown from the ground with a velocity of 20√3 m/s making an angle of 60∘with the horizontal. The ball will be at a height of 40 m from the ground after a time t equal to (g=10 ms−2)

The correct option is C 2secAs s=usinθt−12gt240=20√3×(√3/2)t−12×10×t2⟹5t2−30t+40=0⟹t=2 or 4the minimum time t=2s

A ball is thrown from the ground with a velocity of $20 \sqrt{3}$ m/s making an angle of $60^{\circ}$ with the horizontal. The ball will be at a height of $40 m$ from the ground after a time $t$ equal to ( $g = 10 m s^{- 2}$ )

The correct option is C $2 s e c$
As $s = u sin θ t - \frac{1}{2} g t^{2}$
$40 = 20 \sqrt{3} \times (\sqrt{3} / 2) t - \frac{1}{2} \times 10 \times t^{2}$
$⟹ 5 t^{2} - 30 t + 40 = 0$
$⟹ t = 2 o r 4$
the minimum time $t = 2 s$