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

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:

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

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

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

$1.66 = 0 + 0t + \frac{1}{2}9.8t^2$

$1.66 = 4.9t^2$

$\frac{1.66}{4.9}  = t^2$

$\sqrt{0.339} = t\\ t = 0.582s$

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