Question

When startled, an armadillo will leap upward. Suppose it rises0.544 min first0.200 s, (a) what is its initial speed as it leaves the ground? (b)What is its speed at a height of0.544 m? (c) How much higher does it go?

Short answer

1. Initial speed of an armadillo when it leaves the ground is 3.70 m/s
2. Speed of an armadillo at the height of 0.544 m is 1.74 m/s
3. Armadillo goes higher to 0.154 m

Step-by-step solution

Given information

The given data can be listed below as,

Armadillo rises up to height ${\mathrm{y}}_1=0.544\mathrm{m}$

Armadillo jump in time${\mathrm{t}}_1=0.200\mathrm{s}$

Understanding the kinematic equations

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the second kinematic equation, the initial speed of an armadillo when it leaves the ground can be determined. Further, with the first kinematic equation, the speed of an armadillo at height of 0.544 m can be found. Finally, using the formula for the third kinematic equation, the vertical height of an armadillo can be calculated.

Formula:

The displacement in kinematic equation is given by

$∆y=v_{0}t+\frac{1}{2}a{t}^2$

Determination of initial speed of armadillo

The first kinematic equation for vertical motion is

$∆y=v_{0}t+\frac{1}{2}a{t}^2$

As the upward direction is taken as positive,

localid="1660879109444" $$\begin{gathered} y_1-y_0=\left(v_{0}t\right)-\frac{1}{2}g{t}^2 \\ {v}_0=\frac{∆y+g{t}^2}{2t} \end{gathered}$$

Determination of speed of armadillo at height 0.544m

$$\begin{gathered} {\mathrm{y}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at} \\ \mathrm{v}={\mathrm{v}}_0-\mathrm{gt} \\ \mathrm{v}=3.70\mathrm{m}/{\mathrm{s}}^2-\left(9.8\mathrm{m}/{\mathrm{s}}^2\times 0.200\mathrm{s}\right) \\ \mathrm{v}=1.74\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of armadillo at height of 0.544 m is 1.74 m/s

Determination of speed of armadillo at height 0.544m

$v_f^2=v_0^2+2ay$

Final velocity would be zero at the maximum height

$$\begin{gathered} 0=v_0^2-2a{y}_2 \\ {y}_2=\frac{{\left(3.7\mathrm{m}/\mathrm{s}\right)}^2}{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)} \\ {y}_2=0.698\mathrm{m} \\ ∆y=0.698\mathrm{m}-0.544 \\ =0.154\mathrm{m} \end{gathered}$$

Therefore, the armadillo goes 0.154 m higher.