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
v 2 = ( v 0 ) 2 + 2 ay , \begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned} v 2 = ( v 0 ) 2 + 2 ay , we have
v 2 = ( v 0 ) 2 + 2 ay v 2 − 2 ay = ( v 0 ) 2 v 0 = v 2 − 2 ay \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} v 2 v 2 − 2 ay v 0 = ( v 0 ) 2 + 2 ay = ( v 0 ) 2 = v 2 − 2 ay Substituting the known values,
v 0 = v 2 − 2 ay v 0 = 0 2 − 2 ( − 9.81 m/s 2 ) ( 2.50 m ) v 0 = 7.00 m/s \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} v 0 v 0 v 0 = v 2 − 2 ay = 0 2 − 2 ( − 9.81 m/s 2 ) ( 2.50 m ) = 7.00 m/s 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
y = v o t + 1 2 a t 2 \text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2 y = v o t + 2 1 a t 2 The initial and final position of the kangaroo will be the same, so y y y 2 .
y = v 0 t + 1 2 a t 2 0 = ( 7.00 m/s ) t + 1 2 ( − 9.81 m/s 2 ) t 2 0 = 7 t − 4.905 t 2 7 t − 4.905 t 2 = 0 t ( 7 − 4.905 t ) = 0 t = 0 or 7 − 4.905 t = 0 \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} y 0 0 7 t − 4.905 t 2 t ( 7 − 4.905 t ) t = 0 = v 0 t + 2 1 a t 2 = ( 7.00 m/s ) t + 2 1 ( − 9.81 m/s 2 ) t 2 = 7 t − 4.905 t 2 = 0 = 0 or 7 − 4.905 t = 0 Discard the time 0 since this refers to the beginning of motion. Therefore, we have
7 − 4.905 t = 0 4.905 t = 7 t = 7 4.905 t = 1.43 s \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} 7 − 4.905 t 4.905 t t t = 0 = 7 = 4.905 7 = 1.43 s The kangaroo is about 1.43 seconds long in the air.