A particle is projected up the incline such that its component of velocity along the incline is 10m/s. Time of flight is 2sec and maximum height above the incline is 5m. Then velocity of projection will be
A
10m/s
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B
10√2m/s
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C
5√5m/s
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D
5√10m/s
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Solution
The correct option is B10√2m/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=u2y−2gcosθh where h is the height above the inclined plane.
At maximum height, vy=0 ⇒0=u2sin2(θ−a)−2gcosa×5⇒u2sin2(θ−a)=10gcosa...(i)
We know that time of flight of a projectile on an inclined plane is given by T=2usin(θ−a)gcosα ⇒2=2usin(θ−α)gcosα⇒usin(θ−α)=gcosα...(ii)
Dividing (i) from (ii), we get usin(θ−α)=10 We already know ucos(θ−α)=10 Squaring and adding, u2sin2(θ−a)+u2cos2(θ−a)=200⇒u=10√2ms−1