Answer: D. 0.29 m
Explanation:
We will use the following equations to describe the leap of the cat:
[tex]y=V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] Â (1)
[tex]V_{y}=V_{oy}-gt[/tex] Â (2)
Where:
[tex]y[/tex] Â is the height of the cat Â
[tex]V_{oy}=V_{o}sin\theta[/tex] is the cat's initial velocity
[tex]\theta=60\°[/tex]
[tex]g=9.8m/s^{2}[/tex] Â is the acceleration due gravity
[tex]t[/tex] is the time
[tex]V_{y}[/tex] is the y-component of the velocity
Now the cat will have its maximum height [tex]y_{max}[/tex] when [tex]V_{y}=0[/tex]. So equation (2) is rewritten as:
[tex]0=V_{oy}-gt[/tex] Â (3)
Finding [tex]t[/tex]:
[tex]t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}[/tex] Â (4)
[tex]t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}[/tex]  (5)
[tex]t=0.24 s[/tex] Â (6)
Substituting (6) in (1):
[tex]y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}[/tex]  (7)
Finally:
[tex]y_{max}=0.287 m \approx 0.29 m[/tex] Â (8)
