First slide

Projection Under uniform Acceleration

Question

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

Moderate

A

$10 \sqrt{3} {ms}^{- 1}$
B

$20 \sqrt{3} {ms}^{- 1}$
C

$10 \sqrt{5} {ms}^{- 1}$
D

$20 \sqrt{2} {ms}^{- 1}$

Solution

After one second it is moving 45⁰ with horizontal
$\tan 45^{0} = \frac{usinθ - g \times 1}{ucosθ} = 1 ucosθ = usinθ - g \Rightarrow \Rightarrow (1) after two seconds it moves horizontally means it is at the higest point \frac{usinθ}{g} = 2 usinθ = 2 g substituting in (1) ucosθ = g u = \sqrt{{(usinθ)}^{2} + {(ucosθ)}^{2}} u = \sqrt{{(2 g)}^{2} + {(g)}^{2}} = \sqrt{5} g = 10 \sqrt{5}$

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