Answer:
[tex]t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2 \cdot 1.2 m}{9.81 m/s^2}}=0.50 s[/tex]
Explanation:
The equation that we can use to calculate the time it takes for the raccoon to fall to the ground is:
[tex]t=\frac{2S}{g}[/tex]
where S=1.2 m is the height of the tree and g=9.81 m/s^2 is the acceleration due to gravity. This equation is derived from the equation of the distance in a uniformly accelerated motion, which is given by
[tex]S=S_0 + v_0 t + \frac{1}{2}at^2[/tex]
where S0 is the initial position, v0 is the initial velocity and t the time. In this problem, we can put S0=0 (we can take the initial position as the initial position of the raccoon) and v0=0 (the raccoon starts from rest), so the equation becomes
[tex]S=\frac{1}{2}at^2[/tex]
and since the motion is a free fall, the acceleration is equal to the acceleration of gravity, so a=g:
[tex]S=\frac{1}{2}gt^2[/tex]
And by re-arranging it, we find
[tex]t=\sqrt{\frac{2S}{g}}[/tex]
By substituting numbers, we find
[tex]t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2 \cdot 1.2 m}{9.81 m/s^2}}=0.50 s[/tex]
