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Question

Trajectory of particle in a projectile motion is given as y = x - x280 .

Here , x and y are in meters. For this projectile motion match the following with g = 10 m / s2

"coloumn1coloumn2(a)Angle of Projection(p)20(b)Angle of velocity with horizontal after 4s(q)80(c)Maximum Hieght(r)45(d)Horizontal Range(s)tan112 "


A

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B

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C

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D

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Solution

The correct option is D


Comparing with the standard equation of projectile,

y = x tan θ - gx22u2cos2θ

We get , θ = 45 and u = 20 2 m/s

Time period of this projectile is 4s . Hence after 4s velocity vector will again make 45 with horizontal


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