Answer:
t = 3.88 seconds
Explanation:
Given that,
A ball is thrown horizontally from the top of a 73.9-m building and lands 183 m from the base of the building.
We need to find the time for which the ball is in air. Let the time be t.
[tex]y=y_o+ut+\dfrac{1}{2}at^2[/tex]
Here, a = -g, [tex]y_o=73.9\ m[/tex] and y = 0
u is initial velocity, u = 0
[tex]y_o=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2y_o}{g}}\\\\t=\sqrt{\dfrac{2\times 73.9}{9.8}}\\\\t=3.88\ s[/tex]
So, the ball is in the air for 3.88 seconds.