Answer: 0.336 m
Explanation:
This situation is a good example of the parabolic motion, in which the travel of the squirrel has two components: x-component and y-component. Being their main equations as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
[tex]V_{x}=V_{o}cos\theta[/tex] (2)
Where:
[tex]V_{o}=4m/s[/tex] is the squirrel's initial speed
[tex]\theta=40\°[/tex] is the angle at which the squirrel jumps into the air
[tex]t[/tex] is the time since the bullet is shot until it hits the ground
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (3)
[tex]V_{y}=V_{o}sin\theta-gt[/tex] (4)
Where:
[tex]y_{o}=0m[/tex] is the initial height of the squirrel
[tex]y=0[/tex] is the final height of the squirrel (when it finally hits the ground)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Now, we have to find the maximum height [tex]y_{max}[/tex] reached by the squirrel, and this happens when [tex]V_{y}=0[/tex]. So equation (2) is rewritten as:
[tex]V_{y}=0=V_{o}sin\theta-gt[/tex] (5)
[tex]gt=V_{o}sin\theta[/tex] (6)
[tex]t=\frac{V_{o}sin\theta}{g}[/tex] (7)
Solving (7):
[tex]t=\frac{(4m/s)(sin 40\°)}{9.8m/s^{2}}[/tex] (8)
[tex]t=0.262s[/tex] (9)
Substituting the value of [tex]t[/tex] (9) in (3):
[tex]y_{max}=0+(4m/s)(sin 40\°)(0.262s) - \frac{9.8m/s^{2}{(0.262s)}^{2}}{2}[/tex] (10)
Finally:
[tex]y_{max}=0.336m[/tex] (11)
