Question

using the time of flight and the height when it exploded, you are going to find the initial and final velocity (speed and direction) of the firework. make sure you put the negative in your answer if that is what you get when you do the math. enter answers dont include units. initial velocity (m/s): blank final velocity (m/s): blank check. explosion height = 222 m. t = 5.60 s

Explanation:

Step1: Use kinematic - equation for height

The kinematic equation for vertical displacement $y - y_0=v_0t-\frac{1}{2}gt^2$, where $y - y_0 = 222$ m, $t = 5.60$ s and $g = 9.8$ m/s². Rearranging for initial velocity $v_0$ gives $v_0=\frac{y - y_0+\frac{1}{2}gt^2}{t}$.
$v_0=\frac{222+\frac{1}{2}\times9.8\times(5.60)^2}{5.60}=\frac{222 + \frac{1}{2}\times9.8\times31.36}{5.60}=\frac{222+153.664}{5.60}=\frac{375.664}{5.60}\approx67.1$ m/s.

Step2: Use kinematic - equation for final velocity

The kinematic equation $v = v_0 - gt$. Substitute $v_0\approx67.1$ m/s, $g = 9.8$ m/s² and $t = 5.60$ s.
$v=67.1-9.8\times5.60=67.1 - 54.88 = 12.22$ m/s.

Answer:

Initial Velocity: 67.1
Final Velocity: 12.22