Answer:
the time taken for the object to fall is 6 s.
Explanation:
Given;
final velocity of the object, v = 58.8 m/s
initial velocity of the object, u = 0
The height of fall of the object is calculated as;
v² = u² + 2gh
v² = 2gh
[tex]h = \frac{v^2}{2g} \\\\h = \frac{(58.8)^2}{2(9.8)} \\\\h = 176.4 \ m[/tex]
The time to fall through the height is calculated as;
[tex]h =ut+ \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t= \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 176.4}{9.8} } \\\\t = 6 \ s[/tex]
Therefore, the time taken for the object to fall is 6 s.
