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?
Part A
The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.
From
\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned}we have
\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\
\text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}}
\end{aligned}Substituting the known values,
\begin{aligned}
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\
\text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\
\text{v}_0&= {\color{green}7.00 \ \text{m/s}}
\end{aligned}Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.
Part B
Since the motion of the kangaroo has uniform acceleration, we can use the formula
\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2The initial and final position of the kangaroo will be the same, so y is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s2.
\begin{aligned}
\text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\
0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\
0 & =7\text{t}-4.905\text{t}^{2}\\
7\text{t}-4.905\text{t}^{2}&=0 \\
\text{t}\left( 7-4.905\text{t} \right) & =0 \\
\text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\
\end{aligned}Discard the time 0 since this refers to the beginning of motion. Therefore, we have
\begin{aligned}
7-4.905\text{t} &=0 \\
4.905\text{t} & = 7 \\
\text{t} & =\frac{7}{4.905} \\
\text{t}&={\color{green}1.43 \ \text{s}}
\end{aligned}The kangaroo is about 1.43 seconds long in the air.