A ball is thrown at an angle $53^{\circ}$ with a velocity of 50 $m s^{- 1}$ . Calculate the difference between the velocities at the two instants when the body is 20 m above the ground

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Solution

The correct option is B
$V_{x} = u c o s 53^{\circ} = 30 m s^{- 1} v_{y}^{2} = {(u s i n θ)}^{2} - 2 g h \Rightarrow v_{y} = \pm 20 \sqrt{3} m s^{- 1} ∴ {\to v}_{1} = 30^i + 20 \sqrt{3}^j a n d {\to v}_{2} = 30^i - 20 \sqrt{3}^j ∴ Δ v = {\to v}_{1} - {\to v}_{2} = 40 \sqrt{3}^j m s^{- 1}$

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A ball is thrown at an angle 53∘ with a velocity of 50 ms−1. Calculate the difference between the velocities at the two instants when the body is 20 m above the ground

The correct option is B Vx=u cos 53∘=30ms−1v2y=(u sinθ)2−2gh⇒vy=±20√3 ms−1∴→v1=30^i+20√3^jand →v2=30^i−20√3^j∴Δv=→v1−→v2=40√3^j ms−1

A ball is thrown at an angle $53^{\circ}$ with a velocity of 50 $m s^{- 1}$ . Calculate the difference between the velocities at the two instants when the body is 20 m above the ground

The correct option is B
$V_{x} = u c o s 53^{\circ} = 30 m s^{- 1} v_{y}^{2} = {(u s i n θ)}^{2} - 2 g h \Rightarrow v_{y} = \pm 20 \sqrt{3} m s^{- 1} ∴ {\to v}_{1} = 30^i + 20 \sqrt{3}^j a n d {\to v}_{2} = 30^i - 20 \sqrt{3}^j ∴ Δ v = {\to v}_{1} - {\to v}_{2} = 40 \sqrt{3}^j m s^{- 1}$