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    Question

    A ball is projected from the ground at 45°. It strikes

    a vertical wall and rebounds with the same speed making the same angle on the other side. If the wall is at a distance R/2, where R is the total range, what is the total time of flight?
    A (2√2u)/g Correct Answer Incorrect Answer
    B u/(√2g) Correct Answer Incorrect Answer
    C √2u/g Correct Answer Incorrect Answer
    D u/g Correct Answer Incorrect Answer

    Solution

    We are given:

    • A ball is projected at 45° and rebounds with same speed and same angle after hitting a vertical wall.
    • The wall is located at R/2, where R is the total horizontal range of the projectile if no wall were present.
    • We are to find the total time of flight after hitting and rebounding.
    For a projectile at angle 45°, total horizontal range is: R = (u²·sin2θ)/g = (u²·sin90°)/g = u²/g So, R = u²/g So the wall is at: x = R/2 = (u²)/(2g) Now, for symmetric projectile motion:
    • At R/2, the ball is at half the horizontal range, so the time to reach wall is T/2, where T is the total time of flight without the wall.
    • Total time of flight without wall:
      T = 2u·sinθ / g = 2u·(1/√2) / g = (2u)/(√2·g) = √2·u/g
    So, time to reach wall = T/2 = (√2·u)/(2g) After hitting the wall:
    • The ball reverses horizontal direction, but keeps same speed and same 45° angle (as per question).
    • So its remaining path is a mirror image of the first half.
    Thus, time from wall to ground again = same as time to wall = (√2·u)/(2g) = (√2·u)/(2g) + (√2·u)/(2g) = (√2·u)/g But from earlier: Total time of flight = T = √2·u/g So, total time of flight remains same as original full projectile flight .

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