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 the equation $h = v_0t-\frac{1}{2}gt^2$, where $h$ is the height, $v_0$ is the initial velocity, $t$ is the time, and $g = 9.8\ m/s^2$. Here, $h = 35.0\ m$, $t = 3.25\ s$, and $g=9.8\ m/s^2$.

Step2: Rearrange the equation for $v_0$

Starting from $h = v_0t-\frac{1}{2}gt^2$, we can rewrite it as $v_0t=h + \frac{1}{2}gt^2$. Then $v_0=\frac{h+\frac{1}{2}gt^2}{t}$.

Step3: Substitute the values

Substitute $h = 35.0\ m$, $t = 3.25\ s$, and $g = 9.8\ m/s^2$ into the formula. First, calculate $\frac{1}{2}gt^2=\frac{1}{2}\times9.8\times(3.25)^2=\frac{1}{2}\times9.8\times10.5625 = 51.75625\ m$. Then $h+\frac{1}{2}gt^2=35.0 + 51.75625=86.75625\ m$. Finally, $v_0=\frac{86.75625}{3.25}\approx26.7\ m/s$.

Answer:

The initial velocity of the stone is approximately $26.7\ m/s$.