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Question

A particle is projected up the incline such that its component of velocity along the incline is 10 m/s. Time of flight is 2 sec and maximum height above the incline is 5 m. Then velocity of projection will be

A
10 m/s
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B
102 m/s
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C
55 m/s
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D
510 m/s
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Solution

The correct option is B 102 m/s

Let the direction along the inclined plane be x and the direction perpendicular to the inclined plane be y.
Let T be the time of flight of the projectile.
Using third equation of motion along y-axis, we get
v2y=u2y2gcosθh
where h is the height above the inclined plane.

At maximum height, vy=0
0=u2sin2(θa)2gcosa×5u2sin2(θa)=10gcosa ...(i)

We know that time of flight of a projectile on an inclined plane is given by
T=2usin(θa)gcosα
2=2usin(θα)gcosαusin(θα)=gcosα ...(ii)

Dividing (i) from (ii), we get
usin(θα)=10
We already know ucos(θα)=10
Squaring and adding,
u2sin2(θa)+u2cos2(θa)=200u=102 ms1

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