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

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 .