A ball is projected from the ground at angle - with horizontal- After 1s- it is moving at angle 45- with horizontal and after 2 s it is moving horizontally- What is velocity of projection of the ball-A- 10 -3 m - sB- 20 -3 m - sC- 10 -5 m - sD- 20 -2 m - s

The correct option is C 10√(5) m/s tanθ=V_y/V_x=u sin θ gt/u cos θ Given α =45^∘ when t=1 and, α=0 when t=2 ucosθ=usingθ g.....(1), 2g=usinθ.....(2) from (1) ...

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Byju's Answer

Standard XII

Physics

Definition of a Projectile

A ball is pro...

Question

A ball is projected from the ground at angle $θ$ with horizontal. After 1s, it is moving at angle $45^{\circ}$ with horizontal and after 2s it is moving horizontally. What is velocity of projection of the ball?

$10 \sqrt{3} m / s$

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$20 \sqrt{3} m / s$

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$10 \sqrt{5} m / s$

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$20 \sqrt{2} m / s$

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Solution

The correct option is C
$10 \sqrt{5} m / s$

$t a n θ = \frac{V_{y}}{V_{x}} = \frac{u s i n θ - g t}{u c o s θ} G i v e n α = 45^{\circ} w h e n t = 1 a n d, α = 0 w h e n t = 2 u c o s θ = u s i n g θ - g . . . . . (1), 2 g = u s i n θ . . . . . (2) from (1) and (2) u s i n θ = 2 g a n d u c o s θ = g u = \sqrt{5} g = 10 \sqrt{5} m / s$

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A ball is projected from the ground at angle θwith horizontal. After 1s, it is moving at angle 45∘ with horizontal and after 2s it is moving horizontally. What is velocity of projection of the ball?

The correct option is C 10√5 m/s tanθ=VyVx=u sin θ−gtu cos θGiven α=45∘ when t=1 and, α=0 when t=2ucosθ=usingθ−g.....(1), 2g=usinθ.....(2)from (1) and (2)usinθ=2g and ucosθ=gu=√5g=10√5 m/s

A ball is projected from the ground at angle $θ$ with horizontal. After 1s, it is moving at angle $45^{\circ}$ with horizontal and after 2s it is moving horizontally. What is velocity of projection of the ball?