Question

formula 1 point a ball is thrown straight up with an initial velocity of 38 m/s. how high will it go? round to 1 decimal place. answer

Explanation:

Step1: Identify the relevant formula

The kinematic - equation $v^{2}=v_{0}^{2}-2gh$ is used. At the maximum - height, the final velocity $v = 0$. The initial velocity $v_{0}=38$ m/s and the acceleration due to gravity $g = 9.8$ m/s².

Step2: Rearrange the formula for height $h$

From $v^{2}=v_{0}^{2}-2gh$, when $v = 0$, we can solve for $h$:
\[

$$\begin{align*} 0&=v_{0}^{2}-2gh\\ 2gh&=v_{0}^{2}\\ h&=\frac{v_{0}^{2}}{2g} \end{align*}$$

\]

Step3: Substitute the values

Substitute $v_{0}=38$ m/s and $g = 9.8$ m/s² into the formula:
\[
h=\frac{38^{2}}{2\times9.8}=\frac{1444}{19.6}\approx73.7
\]

Answer:

73.7