First slide

Projection Under uniform Acceleration

Question

A ball is projected from the ground at angle $θ$ with the horizontal. After 1 s, 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

Suppose the angle made by the instantaneous velocity with the horizontal be $α .$ then
$tanα = \frac{v_{y}}{v_{x}} = \frac{usinθ - gt}{ucosθ} Given that α = 45 °, when t = 1 s; α = 0 °, when t = 2 s . This gives ucosθ = usinθ - g and usinθ - 2 g = 0 Solving Eqs. (i) and (ii), we find usinθ = 2 g and ucosθ = g . Squaring and adding, u = \sqrt{5} g = 10 \sqrt{5} {ms}^{- 1}$

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