Question

1. -/1 points a mountain lion jumps to a height of 2.70 m when leaving the ground at an angle of 40.0°. what is its initial speed (in m/s) as it leaves the ground? m/s resources read it

Explanation:

Step1: Analyze vertical - motion

The vertical - component of the initial velocity is $v_{0y}=v_0\sin\theta$, where $v_0$ is the initial speed and $\theta = 40.0^{\circ}$. At the maximum height $h = 2.70\ m$, the vertical velocity $v_y = 0$. We use the kinematic equation $v_y^2=v_{0y}^2 - 2gh$.

Step2: Substitute $v_{0y}$ into kinematic equation

Since $v_y = 0$, the equation becomes $0=(v_0\sin\theta)^2-2gh$. We can solve this equation for $v_0$. First, we can rewrite it as $(v_0\sin\theta)^2 = 2gh$. Then $v_0^2=\frac{2gh}{\sin^{2}\theta}$.

Step3: Calculate $v_0$

Given $g = 9.8\ m/s^{2}$, $h = 2.70\ m$, and $\theta = 40.0^{\circ}$, we have $\sin\theta=\sin40.0^{\circ}\approx0.6428$. Then $v_0=\sqrt{\frac{2gh}{\sin^{2}\theta}}=\sqrt{\frac{2\times9.8\times2.70}{0.6428^{2}}}$.
$v_0=\sqrt{\frac{52.92}{0.4132}}\approx\sqrt{128.07}\approx11.3\ m/s$.

Answer:

$11.3$