Question

A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the highest point of the cat’s trajectory?

A. 0.19 m


B. 10.96 m


C. 0.58 m


D. 0.29 m

Answer 1

Answer: D. 0.29 m

Step-by-step explanation:

We will use the following equations to describe the leap of the cat:

`y=V_(o)sin\theta t-(gt^(2))/(2)` (1)

`V_(y)=V_(oy)-gt` (2)

Where:

`y` is the height of the cat

`V_(oy)=V_(o)sin\theta` is the cat's initial velocity

`\theta=60\°`

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

`t` is the time

`V_(y)` is the y-component of the velocity

Now the cat will have its maximum height
`y_(max)` when
`V_(y)=0`. So equation (2) is rewritten as:

`0=V_(oy)-gt` (3)

Finding
`t`:

`t=(V_(oy))/(g)=(V_(o)sin\theta)/(g)` (4)

`t=(2.74 m/s sin(60\°))/(9.8m/s^(2))` (5)

`t=0.24 s` (6)

Substituting (6) in (1):

`y_(max)=(2.74 m/s)sin(60\°) (0.24 s)-((9.8m/s^(2))(0.24 s)^(2))/(2)` (7)

Finally:

`y_(max)=0.287 m \approx 0.29 m` (8)