Question

A squirrel jumps into the air with a velocity of 4m/s at an angle of 40°. What is the maximum height reached by the squirrel?

Answer 1

0.34 m

Step-by-step explanation:

Apex ; good luck yall! :)

Answer 2

Answer: 0.336 m

Step-by-step 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:

`x=V_(o)cos\theta t` (1)

`V_(x)=V_(o)cos\theta` (2)

Where:

`V_(o)=4m/s` is the squirrel's initial speed

`\theta=40\°` is the angle at which the squirrel jumps into the air

`t` is the time since the bullet is shot until it hits the ground

y-component:

`y=y_(o)+V_(o)sin\theta t-(gt^(2))/(2)` (3)

`V_(y)=V_(o)sin\theta-gt` (4)

Where:

`y_(o)=0m` is the initial height of the squirrel

`y=0` is the final height of the squirrel (when it finally hits the ground)

`g=9.8m/s^(2)` is the acceleration due gravity

Now, we have to find the maximum height
`y_(max)` reached by the squirrel, and this happens when
`V_(y)=0`. So equation (2) is rewritten as:

`V_(y)=0=V_(o)sin\theta-gt` (5)

`gt=V_(o)sin\theta` (6)

`t=(V_(o)sin\theta)/(g)` (7)

Solving (7):

`t=((4m/s)(sin 40\°))/(9.8m/s^(2))` (8)

`t=0.262s` (9)

Substituting the value of
`t` (9) in (3):

`y_(max)=0+(4m/s)(sin 40\°)(0.262s) - \frac{9.8m/s^(2){(0.262s)}^(2)}{2}` (10)

Finally:

`y_(max)=0.336m` (11)