Answer:
t = 2.44 s
Explanation:
Given that,
A stunt man climb up 29.4 meters into a cannon. He gets fired horizontally out of the cannon with a speed of 57.1 m/s.
We need to find the time for which was the stunt man in the air. Let it is t. It can be calculated using second equation of motion as follows :
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
u is initial speed and it is 0
[tex]s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 29.4}{9.8}} \\\\t=2.44\ s[/tex]
So, the stunt man is in the air for 2.44 seconds.
