A kangaroo jumps straight up to a vertical height of 1.19m, how long was it in the air before returning on earth

why did you multiply it by 2

To determine how long the kangaroo was in the air before returning to the Earth, we can use the equation of motion for a freely falling object.

The equation is:
H = (1/2) * g * t^2

Where:
H is the height (1.19m in this case),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time in seconds.

First, let's rearrange the equation to solve for time (t):

t^2 = (2H / g)

t = sqrt((2H / g))

Plugging in the values:

t = sqrt((2 * 1.19) / 9.8)

t = sqrt(0.242)

t ≈ 0.49 seconds

Therefore, the kangaroo was in the air for approximately 0.49 seconds before returning to the Earth.

To find the time the kangaroo was in the air, we can use the equation of motion for vertical motion:

h = v₀t + (1/2)gt²

where:
- h is the vertical height (1.19m in this case),
- v₀ is the initial vertical velocity (which is zero since the kangaroo jumps straight up),
- g is the acceleration due to gravity (-9.8 m/s² for Earth's gravity),
- t is the time.

Since the initial velocity is zero, our equation becomes:

h = (1/2)gt²

Rearranging the equation to solve for t:

t² = (2h) / g

t = √[(2h) / g]

Now, let's substitute the given values into the equation:

t = √[(2 * 1.19) / 9.8]

t ≈ √0.243

t ≈ 0.493 seconds

Therefore, the kangaroo was in the air approximately for 0.493 seconds before returning to Earth.

h = 0.5g*t^2 = 1.19 m.

4.9t^2 = 1.19
t^2 = 0.243
Tf = 0.493 s. = Fall time.

Tr = Tf = 0.493 s. = Rise time.

Tr+Tf = 0.493 + 0.493 = 0.986 s. = Time
in air.