Question

trial 2:
recorded data:
v = 10 m/s
t=

s
m (recorded

h=
from simulation)
calculate maximum height h, show all work and units:

Explanation:

Step1: Identify the relevant kinematic - equation

At maximum height, the final vertical velocity $v_y = 0$. The kinematic equation $v_y^2=v_{0y}^2 - 2gh$ (where $v_{0y}$ is the initial vertical velocity, $g = 9.8\ m/s^2$ is the acceleration due to gravity and $h$ is the height) can be used. If the initial velocity $v = 10\ m/s$ is assumed to be vertical (no angle given, so we assume vertical motion for maximum - height calculation), $v_{0y}=10\ m/s$.

Step2: Rearrange the kinematic equation to solve for $h$

Starting with $v_y^2=v_{0y}^2 - 2gh$, when $v_y = 0$, we can rewrite it as $h=\frac{v_{0y}^2}{2g}$.

Step3: Substitute the values

Substitute $v_{0y}=10\ m/s$ and $g = 9.8\ m/s^2$ into the formula. $h=\frac{(10\ m/s)^2}{2\times9.8\ m/s^2}=\frac{100\ m^2/s^2}{19.6\ m/s^2}\approx5.1\ m$.

Answer:

$h\approx5.1\ m$