Question

A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your answer to three significant figures and include the appropriate units

Answer 1

The kangaroo was 1.164s in the air before returning to Earth

Step-by-step explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

`x = x_(0) + V_(0)t + (1)/(2)at^2`

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

`x = x_(0) + V_(0)t + (1)/(2)at^2`

`1.66 = 0 + 0t + (1)/(2)9.8t^2`

`1.66 = 4.9t^2`

`(1.66)/(4.9)  = t^2`

`√(0.339) = t\\ t = 0.582s`

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