Answer:
The kangaroo was 1.164s in the air before returning to Earth
Explanation:
For this we are going to use the equation of distance for an uniformly accelerated movement, that is:
[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]
Where:
x = Final distance
xo = Initial point
Vo = Initial velocity
a = Acceleration
t = time
We have the following values:
x = 1.66m
xo = 0m (the kangaroo starts from the floor)
Vo = 0 m/s (each jump starts from the floor and from a resting position)
a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)
t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.
Now replace the values in the equation
[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]
[tex]1.66 = 0 + 0t + \frac{1}{2}9.8t^2[/tex]
[tex]1.66 = 4.9t^2[/tex]
[tex]\frac{1.66}{4.9} = t^2[/tex]
[tex]\sqrt{0.339} = t\\ t = 0.582s[/tex]
It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is
t = 0.582s + 0.582s
t = 1.164s
The kangaroo was 1.164s in the air before returning to Earth
