Answer:
The horizontal distance traveled by the ball is 29.7 m.
Explanation:
Given;
height of the building, h = 24.7 m
initial horizontal velocity, Vā = 13.2 m/s
The time taken for the ball to fall from the vertical height is calculated as;
[tex]h = V_yt + \frac{1}{2} gt^2[/tex]
where;
[tex]V_y[/tex] is initial vertical velocity = 0
[tex]h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 24.7}{9.8} } \\\\t = 2.25 \ s[/tex]
The horizontal distance traveled by the ball is calculated as;
[tex]X = V_x t\\\\X = 13.2 \times2.25\\\\X = 29.7 m[/tex]
Therefore, the horizontal distance traveled by the ball is 29.7 m.
