A cat leaps to try to catch a bird.If the cat's jump was at 60° off the ground and its initial velocity was $2.74 \mathrm {~m} / \mathrm { s }$ what is the highest point of its trajectory,neglecting effects of air?
A)0.29 m
B)0.58 m
C)10.96 m
D)0.19 m

Question

Quizplus · Accepted Answer

The highest point of the trajectory, or the maximum height $h$, can be calculated using the formula $h = \frac{v^2 \sin^2(\theta)}{2g}$, where $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity (approximately $9.81 \mathrm{m/s^2}$). Plugging in the given values: $h = \frac{(2.74)^2 \sin^2(60)}{2 \times 9.81}$, we find that $h \approx 0.29 \mathrm{m}$.

A Cat Leaps to Try to Catch a Bird 2.74 m/s2.74 \mathrm {~m} / \mathrm { s }2.74 m/s

A Cat Leaps to Try to Catch a Bird $2.74 \mathrm {~m} / \mathrm { s }$