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Look at the picture...If you through the projectile horizontally, its vertical component is zero. It land on the final spot without bouncing.

The time taken to fall 0.9 m is 0.429s.

Horizontal velocity to cover 2.5 m in 0.429 s is

Vo = 2.5/0.429s = 5.833 m/s.

To land on the same final spot, the other velocities with different angles of projection are such that their horizontal component must be 5.833 and time of flight must be 0.429. Because of the larger velocity it will move larger distance because of bouncing.

In the given problem the angle of projection is 45 degree. So the initial velocity will be such that vi*sin45 = 5.833. Find Vi.

To find the time for the ball to fall 1 meter where it bounces I use t=root(2h/g) and I get t=0.452s. To find the velocity when the ball bounces I use v=root(2gh) and I get v=4.43m/s. I now use that velocity with yf=yi+vt+1/2gt^2 to solve for t where yf=0.1m and yi=0m. This gives me t=0.881s. Adding the two times together I get 0.452s+0.881s=1.333s. Now to find the velocity I use v=d/t where d is 2.5m and get the initial velocity V=1.875m/s. That is in the horizontal direction. So assuming the ball was shot at 0 degrees instead of 45 that would be the answer. Right?