Question

Squirrel completing a short glide travels in a straight line tipped 40 â below the horizontal. the squirrel starts 9.0 m above the ground on one tree and glides to a second tree that is a horizontal distance of 3.5 m away. what is the length of the squirrel's glide path

Answer 1

The distance travel by the squirrel along the gliding path is $\boxed{4.57{\text{ m}}}$.

Further explanation:

Here, we have to calculate the actual distance travel by the squirrel along its gliding path.

As given that, when squirrel travel its path of the gliding is at $40^\circ$ with the horizontal as like angle of the depression.

The squirrel is at the height of $9{\text{ m}}$ from the ground on the first tree and the position of the second tree is at the distance of $3.5{\text{ m}}$ from the first tree.

Now, the squirrel glides from the first tree to the second tree not in the horizontal direction but at an angle of $40^\circ$  in clockwise downward direction with the horizontal.

First of all we will draw the diagram as shown in figure attached below.

From figure,

The actual distance travel by the squirrel is $AC$.

So, in $\Delta ABC$, apply cosine rule,

$\cos 40^\circ=\dfrac{{AB}}{{AC}}$

Here, $AB$ is the distance between the first and second tree.

From above equation,

$AC = \dfrac{{AB}}{{\cos 40^\circ }}$

Substitute the value of the $AB$ as $3.5{\text{ m}}$ in the above equation.

$\begin{aligned}AC&=\dfrac{{3.5}}{{\cos 40^\circ }}\\&=4.57{\text{ m}}\\\end{aligned}$

Thus, the distance travel by the squirrel along the gliding path between the first and second tress is $\boxed{4.57{\text{ m}}}$.

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Answer detail:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Squirrel, short glide, below horizontal, above the ground, jumps, one tree, second tree, glides, 3.5m away, glide path, 9.0m.

Answer 2

The length of squirrel's glide path is 4.6 m.

The squirrel starts from the point A of the tree 1 and glides along a straight line path AC at an angle 40° to the horizontal. It lands on another tree situated at a horizontal distance of 3.5 m from the first tree.

Refer the diagram attached.

In the ΔABC, ∠ACB =40° ( alternate interior angles)

$cos(ACB)=\frac{BC}{AC}$

Therefore,

$AC=\frac{BC}{cos(ACB)}$

Substitute 3.5 m for BC and 40°for ∠ACB.

$AC=\frac{BC}{cos(ACB)}\\ =\frac{3.5 m}{cos40^o} \\ =4.5689m$

The length of the squirrel's glide path is 4.6 m