Answer :
Given data:
Initial speed;
[tex]u=15\text{ m/s}[/tex]Height;
[tex]H=1.5\text{ m}[/tex]The time of flight for horizontal projectile i s given as,
[tex]T=\sqrt[]{\frac{2H}{g}}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} T=\sqrt[]{\frac{2\times1.5\text{ m}}{9.8m/s^2}} \\ =0.55\text{ s} \end{gathered}[/tex]Therefore, the ball will take 0.55 s to land. Hence, option (a) is the correct choice.
The horizontal range is given as,
[tex]R=u\sqrt[]{\frac{2H}{g}}[/tex]Substituting all known values,
[tex]\begin{gathered} R=(15\text{ m/s})\times\sqrt[]{\frac{2\times1.5\text{ m}}{9.8\text{ m/s}^2}} \\ =8.29\text{ m} \\ \approx8.3\text{ m} \end{gathered}[/tex]Therefore, the ball will cover a distance of 8.3 m. Hence, option (c) is the correct choice.