We are given that a firework is launched from the ground and it is required that it reaches a height of 240m. Since we are ignoring the drag we can use the equation of motion of an object in free fall. Specifically, we will use the formula for the maximum height of an object:
[tex]H=\frac{v^2_0}{2g}[/tex]Where:
[tex]\begin{gathered} H\text{ = height} \\ v_0=\text{ initial sp}ed \\ g\text{ = acceleration of gravity} \end{gathered}[/tex]We solve for the initial speed first by multiplying both sides by "2g":
[tex]2gH=v^2_0[/tex]Now we take the square root to both sides:
[tex]\sqrt[]{2gH}=v_0[/tex]Now we replace the given values:
[tex]\sqrt[]{2(9.8\frac{m}{s^2})(240m)}=v_0[/tex]Solving the operations:
[tex]68.6\frac{m}{s}=v_0[/tex]Since we are asked to determine the speed in miles per hour we will use the following conversion factors:
[tex]\begin{gathered} 1\text{ mile = }1609.34\text{ m}eters \\ 1\text{ hour = 3600 seconds} \end{gathered}[/tex]Now we multiply the speed by the conversion factors in the form of fractions:
[tex]68.6\frac{m}{s}\times\frac{1\text{mile}}{1609.34m}\times\frac{3600s}{1h}[/tex]Solving the operation:
[tex]v_0=153.42\text{ mph}[/tex]Therefore, the launch speed must be 153.42 mph.
If the velocity is 160 miles per hour it would be safe because a larger speed would increase the maximum height Since the height is proportional to the square of the speed.