A kangaroo jumps up with an initial velocity of 36 feet per second from the ground (assume
its starting height is O feet). Use the vertical
motion model, h = -1612 + vt + s, where y
is the initial velocity in feet per second and s
is the height in feet, to calculate the amount of
time the kangaroo is in the air before it hits the
ground again. Round your answer to the
nearest tenth if necessary.
Time in air seconds
Enter the answer

The vertical launch of kinematics allows to find the flight time of the kangaroo is

             t = 2.2 s

Given parameters

To find

Vertical launch is an application of kinematics where there is no motion on the ax axis and on the y axis the acceleration is the acceleration of gravity (g = 32.16 ft / s²), the equation of motion is

               y = y₀ + v₀ t - ½ g t²

The expression written in the exercise has some typing errors.  Where y and y₀ are the current and initial height, respectively, v₀ the initial velocity, g the acceleration of gravity and t the time

The time of flight occurs when the kangaroo reaches the ground (y = 0)

              0 = 0 + v₀ t - ½ g t²

              t (v₀ - ½ g t) = 0

This expression has two solutions

             t = 0 s

This solution corresponds to the time of departure and the solution for the time of arrival on the ground is

             t =

             t = 2 36 / 32.16

            t = 2.24 s

Using the proposed criterion of leaving a single decimal the answer is

              t = 2.2 s

In conclusion with the vertical launch of kinematics we can find the flight time of the kangaroo is

             t = 2.2 s

learn more about vertical launch here:

link

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.