question 22\nneglecting air resistance, what maximum height will be reached by an arrow launched straight…

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v^{2}=v_{0}^{2}-2gh$. At the maximum - height, the final velocity $v = 0$. The initial velocity $v_{0}=35\ m/s$ and the acceleration due to gravity $g = 9.8\ m/s^{2}$. $0 = v_{0}^{2}-2gh$

Step2: Solve for the height $h$

Rearrange the equation $0 = v_{0}^{2}-2gh$ to solve for $h$. We get $h=\frac{v_{0}^{2}}{2g}$. Substitute $v_{0}=35\ m/s$ and $g = 9.8\ m/s^{2}$ into the formula: $h=\frac{35^{2}}{2\times9.8}=\frac{1225}{19.6}=62.5\approx63\ m$

Answer:

$63\ m$ (corresponding to the second option)