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

Explanation:

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

$y=V_{o}sin\theta t-\frac{gt^{2}}{2}$   (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 $t$:

$t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}$   (4)

$t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}$   (5)

$t=0.24 s$   (6)

Substituting (6) in (1):

$y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}$   (7)

Finally:

$y_{max}=0.287 m \approx 0.29 m$   (8)