a kangaroo jumps at 6.32 m/s at a 55.8° angle to the horizon. how long does it take for the kangaroo to…

Explanation:

Step1: Find vertical - initial velocity

The initial velocity $v_0 = 6.32$ m/s and the launch angle $\theta=55.8^{\circ}$. The vertical - component of the initial velocity is given by $v_{0y}=v_0\sin\theta$. So, $v_{0y}=6.32\sin(55.8^{\circ})$. $v_{0y}=6.32\times0.827 = 5.237$ m/s.

Step2: Use kinematic equation for vertical motion

At the maximum height, the vertical velocity $v_y = 0$. The kinematic equation $v_y=v_{0y}-gt$ (where $g = 9.8$ m/s²) is used to find the time $t$. Rearranging for $t$ gives $t=\frac{v_{0y}-v_y}{g}$. Since $v_y = 0$, then $t=\frac{v_{0y}}{g}$. Substituting $v_{0y}=5.237$ m/s and $g = 9.8$ m/s², we get $t=\frac{5.237}{9.8}=0.534$ s.

Answer:

$0.534$