Question

During World War I, mortars were fired from trenches 3 feet below ground level. The mortars had a velocity of150 ft/sec. Determine how long it will take for the mortar shell to strike its target.• What is the initial height of the rocket? -3 ft.• What is the maximum height of the rocket? 348.56 ft• How long does it take the rocket to reach the maximum height ? 4.68750 sec.• How long does it take the rocket to hit the ground (ground level)? 9.35 sec.• How long does it take the rocket to hit a one hundred feet tall building that is in it's downward path?[ Select]• What is the equation that represents the path of the rocket? Select]

Answer 1

$\begin{gathered} y=-\frac{g}{2}t^2+v_0t+y_0 \\ g\text{ in feets per seconds is 32} \end{gathered}$$\begin{gathered} \text{Now the height of the bilding is 100 hence y must be 100, i.e., y=100, hence one has} \\ 100=-16t^2+150t-3 \\ or \\ -16t^2+150t-103=0 \\ \text{the solutions are given by:} \end{gathered}$$\begin{gathered} t_1=\frac{-150+\sqrt[]{150^2-4(-16)(-103)}}{2(-16)} \\ t_2=\frac{-150-\sqrt[]{150^2-4(-16)(-103)}}{2(-16)} \end{gathered}$$\begin{gathered} t_1=\frac{-150+\sqrt[]{22500-6592}}{-32} \\ t_1=\frac{-150+\sqrt[]{15908}}{-32} \\ t_1=\frac{-150+126}{-32} \\ t_1=\frac{24}{-32}\text{ This solution is negative, it doesnt work. Let us s}ee\text{ the other solution:} \end{gathered}$$\begin{gathered} t_2=\frac{-150-\sqrt[]{150^2-4(-16)(-103)}}{2(-16)} \\ t_2=\frac{-150-\sqrt[]{22500^{}-6592}}{-32} \\ t_2=\frac{-150-\sqrt[]{15908}}{-32} \\ t_2=\frac{-150-126}{-32} \\ t_2=\frac{-276}{-32} \\ t=\frac{276}{32} \\ t=8.6\text{seg} \\ It\text{ takes 8.6 second to hit the bilding} \end{gathered}$$\begin{gathered} \text{the general equation of the parabolic motion is }y=-\frac{g}{2}t^2+v_0t+y_0\text{. In this case, this is} \\ y=-16t^2+150t-3 \end{gathered}$