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Question

# It is given that the ball hits a wall at a horizontal distance 15m at a point 5m above the point of projection. If a second wall of height (5−dx) m with dx→0 has to be constructed such that the ball just misses to hit this wall, horizontal distance of this wall from the point of projection will be

A
7.5 m
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B
12 m
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C
10.5 m
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D
6.5 m
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Solution

## The correct option is A 7.5 mIn a projectile the particle follow a specific path which call the trajectory of the particle , the equation of the trajectory as follows - y=xtanθ−gx22(ucosθ)2where x is the horizontal displacement and y is the vertical displacement of the particle at any point of the time during projectile. now put the given values in the equation to get the velocity ⟹5=15tan(45∘)−10×1522(ucos(45∘))2⟹u=15m/sAs we know that the particle passes the 5m twice so by again putting all the values in the trajectory equation you will get the another distance,So,⟹5=x×tan(45∘)−10×x22(15×cos(45∘))2⟹x=7.5m So the Horizontal distance of the new wall is 7.5 m.

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