Question

1. (30 points) a slingshot fires a stone straight upward. it takes 3.25 seconds to reach a height of 35.0 meters. with what velocity was the stone fired?

Explanation:

Step1: Identify the kinematic - equation

We use \(h = v_0t-\frac{1}{2}gt^{2}\) for vertical - motion.

Step2: Rearrange the equation for \(v_0\)

We get \(v_0=\frac{h+\frac{1}{2}gt^{2}}{t}\) by algebraic manipulation.

Step3: Substitute the values

Substitute \(h = 35.0\ m\), \(t = 3.25\ s\) and \(g = 9.8\ m/s^{2}\) into the formula for \(v_0\).

Step4: Calculate \(v_0\)

Perform the arithmetic operations to find \(v_0\approx26.7\ m/s\).

Answer:

We can use the kinematic equation \(h = v_0t-\frac{1}{2}gt^{2}\), where \(h = 35.0\ m\), \(t = 3.25\ s\) and \(g= 9.8\ m/s^{2}\), and we need to solve for \(v_0\).

First, rewrite the equation for \(v_0\):

\[

$$\begin{align*} h&=v_0t-\frac{1}{2}gt^{2}\\ v_0t&=h + \frac{1}{2}gt^{2}\\ v_0&=\frac{h+\frac{1}{2}gt^{2}}{t} \end{align*}$$

\]

Substitute \(h = 35.0\ m\), \(t = 3.25\ s\) and \(g = 9.8\ m/s^{2}\) into the formula:

\[

$$\begin{align*} v_0&=\frac{35.0+\frac{1}{2}\times9.8\times(3.25)^{2}}{3.25}\\ &=\frac{35.0 + \frac{1}{2}\times9.8\times10.5625}{3.25}\\ &=\frac{35.0+51.75625}{3.25}\\ &=\frac{86.75625}{3.25}\\ &\approx26.7\ m/s \end{align*}$$

\]