A ball is projected from a point on the floor at a speed of 15 m/s...

The horizontal range of the projectile is given by $R = \dfrac{u^2 \sin 2\theta}{g}$.

Substituting the given values $u = 15 \text{ m/s}$, \theta = 60^{}^{{}^{\circ}}, and $g = 9.8 \text{ m/s}^2$:

R = \dfrac{(15)^2 \sin(120^{}^{{}^{\circ}})}{9.8} = \dfrac{225 \times \sqrt{3}}{2 \times 9.8} \approx 19.88 \text{ m}

Since the distance to the first wall is $5 \text{ m}$, which is less than the range ($5 \text{ m} < 19.88 \text{ m}$), the ball will hit the wall before hitting the floor. Thus, the answer for the $5 \text{ m}$ wall is Yes.

For the second wall at $22 \text{ m}$, the distance is greater than the range ($22 \text{ m} > 19.88 \text{ m}$). The ball will hit the floor before reaching this wall. Thus, it will not hit the $22 \text{ m}$ wall without hitting the floor first.

Therefore, it will hit the $5 \text{ m}$ wall, and the answer differs for the $22 \text{ m}$ wall.