InterviewSolution
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InterviewSolution
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A projectile si thrown with velocity of `50m//s` towards an inclined plane from ground such that is strikes the inclined plane perpendiclularly. The angle of projection of the projectile is `53^(@)` with the horizontal and the inclined plane is inclined at an angle of `45^(@)` to the horizonta.. (a) Find the time of flight. (b) Find the distance between the point of projection and the foot of inclined plane. |
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Answer» Correct Answer - (a) `t = 7 sec`, (b) `175 m` `v_(x) = u_(x) = 50 cos 53^(@) = 30m//s` at striking point `v_(y) =- u_(x) =- 30 m//s` (a) form time of flight `v_(y) = u_(y) +a_(y) t rArr -30 = 40 +(-10)t rArr t = 7 sec` ltbrltgt (b) for distance distance `= OQ - MQ = u_(x)t - (u_(y)t +(1)/(2)a_(y)t^(2))` distance `= 30 xx 7- [40 xx 7 +(1)/(2) xx (-10) xx 7^(2)]` distance `= 210 - (280 - 245) rArr distance `= 175 m` |
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