Q50PE - College Physics (Urone)

Question

A kangaroo can jump over an object 2.50 m high. 

(a) Calculate its vertical speed when it leaves the ground. 

(b) How long is it in the air?

Step-by-Step Solution

Answer

a) $7 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

b) 1.43 s

Step 1: Given data

• Final velocity: V= 0.
• Distance: D= 2.50 m.
• Acceleration: g=-9.81 m/s².

Step 2: Vertical speed or initial velocity of kangaroo

a)

The initial velocity of the kangaroo can be calculated by the following equation:

${\mathbf{V}}^{\mathbf{2}}\mathbf{-}{\mathbf{U}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{gd}$

Here V is the final velocity, g is the acceleration, U is the initial velocity, and d is the distance traveled to achieve the final velocity.

Substituting the values in the above expression, we get:

$\begin{array}{rcl}{\left(0\right)}^2-{\left(\mathrm{U}\right)}^2& =& 2\times \left(-9.81\right)\times \left(2.5\right)\\ \mathrm{U}& =& \sqrt{49.05} \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\\ \mathrm{U}& =& 7 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\end{array}$

 

Hence the initial velocity of the kangaroo is $7 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step 3: Time taken by a kangaroo in the air

b)

Hence, the given data:

• U= $7 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.
• V= $-7 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .
• g = -9.81 m/s²

The time can be calculated from the equation of motion as:

$\mathbf{V}\mathbf{=}\mathbf{U}\mathbf{+}\mathbf{gt}$

 Here t is the time.

Substituting the values in the above expression, we get:

$\begin{array}{rcl}-7& =& 7+\left(-9.81\right)\times \left(t\right)\\ t& =& \frac{-14}{-9.81}\\ & =& 1.43 s\end{array}$

 

The total time spent by the kangaroo in the air is 1.43 s .