Question

question 5 of 10
a squirrel jumps into the air with a velocity of 7 m/s at an angle of 20°. what is the maximum height reached by the squirrel?
a. 0.3 m
b. 0.5 m
c. 0.9 m
d. 0.7 m

Explanation:

Step1: Find vertical - initial velocity

The initial velocity is $v_0 = 7$ m/s and the angle is $\theta=20^{\circ}$. The vertical - initial velocity $v_{0y}$ is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=7\sin20^{\circ}\approx7\times0.342 = 2.394$ m/s.

Step2: Use the kinematic equation for vertical motion

At the maximum height, the final vertical velocity $v_y = 0$. The kinematic equation $v_y^{2}-v_{0y}^{2}=-2gh$ (where $g = 9.8$ m/s²) can be used to find the maximum height $h$. Rearranging the equation for $h$, we get $h=\frac{v_{0y}^{2}-v_y^{2}}{2g}$. Substituting $v_y = 0$ and $v_{0y}\approx2.394$ m/s and $g = 9.8$ m/s², we have $h=\frac{(2.394)^{2}-0^{2}}{2\times9.8}=\frac{5.731}{19.6}\approx0.3$ m.

Answer:

A. 0.3 m