Question

A hawk is flying horizontally at $\mathbf{10}\mathbf{}\mathbf{\text{m/s}}$in a straight$\mathbf{\text{M}}$line $\mathbf{\text{200m}}$,above the ground. A mouse it has been carrying struggles free from its talons. The hawk continues on its path at the same speed for$\mathbf{\text{2.00s}}$before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse$\mathbf{\text{3.00m}}$above the ground. (a) Assuming no air resistance acts on the mouse, find the diving speed of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For what time interval did the mouse experience free fall?

Short answer

(a) The diving speed of the hawk is$46.5\text{m/s}$.

(b) Angle the hawk made with the horizontal while its descent is$77.6^{\circ }$.

(c) Time interval the mouse experience free fall$6.34\text{s}$.

Step-by-step solution

Step 1: The time taken and angle made.

The time taken by the mouse to cover the distance,$\mathit{t}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{h}}{\mathbf{g}}}$

Angle made isrole="math" localid="1663634325350" $\mathit{\theta }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{v}}_{\mathbf{y}}}{{\mathbf{v}}_{\mathbf{x}}}\mathbf{\right)}$

Here ${\mathbf{v}}_{\mathbf{y}}$is the vertical velocity of the hawk,${\mathbf{v}}_{\mathbf{x}}$ is the horizontal velocity of the hawk .

Speed of the hawk.

(a)

Since there is no air resistance, the horizontal velocity of hawk and mouse is same. The total vertical distance travelled by the mouse is

$\begin{array}{c}h=\left(200\mathrm{m}\right)-\left(3\mathrm{m}\right)\\ =197\mathrm{m}\end{array}$

The downward acceleration of the mouse is

$g=9.81\mathrm{m}/{\mathrm{s}}^2$

Hence, the time taken by the mouse to cover the distance is

$\begin{array}{c}t=\sqrt{\frac{2h}{g}}\\ =\sqrt{\frac{2\left(197\mathrm{m}\right)}{\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}}\\ =6.34\mathrm{s}\end{array}$

Hawk started diving after $2\text{s}$. Hence, the hawk covered the distance in $(6.34\mathrm{s})-(2.00\mathrm{s})=4.34\mathrm{s}$.

Since, the hawk moved with constant velocity, the downward velocity of the hawk was

$\begin{array}{c}{v}_y=\frac{197\mathrm{m}}{4.34\mathrm{s}}\\ =45.4\mathrm{m}/\mathrm{s}\end{array}$

The horizontal velocity of the hawk is

$v_x=10.0\mathrm{m}/\mathrm{s}$

Hence, the speed of the hawk is

$\begin{array}{c}v=\sqrt{{(10.0\mathrm{m}/\mathrm{s})}^2+{(45.4\mathrm{m}/\mathrm{s})}^2}\\ =46.5\mathrm{m}/\mathrm{s}\end{array}$

The diving speed of the hawk is $46.5\mathrm{m}/\mathrm{s}$.

Angle made by the hawk.

(b)

Angle made by the hawk is

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)\\ ={\tan }^{-1}\left(\frac{45.4\mathrm{m}/\mathrm{s}}{10.0\mathrm{m}/\mathrm{s}}\right)\\ =77.6°\end{array}$

Therefore, the angle made by the hawk with the horizontal is $77.6°$.

Find the time interval for which mouse experience free fall.

(c)

As calculated in part (a) of the problem, the time taken by mouse to fall from $200\text{m}$to $3\text{m}$is $6.34\text{s}$. Hence, the mouse experienced the free fall for time $6.34\text{s}$.