Hello!
First, let's write some important information contained in the exercise:
• t = 0 (launching)
,• h(t) = -4.9t² +250t +122
We can solve it as a quadratic equation, equaling it to 0:
[tex]-4.9t^{2}+250t+122=0[/tex]Let's find the coefficients:
• a = -4.9
,• b = 250
,• c = 122
Now, let's replace the values in the quadratic formula:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2\cdot a} \\ \\ x=\frac{-250\pm\sqrt[]{250^2-4\cdot(-4.9)\cdot122}}{2\cdot(-4.9)} \\ \\ x=\frac{-250\pm\sqrt[]{62500^{}+2391.2}}{-9.8} \\ \\ x=\frac{-250\pm\sqrt[]{64891.2}}{-9.8} \\ \\ x=\frac{-250\pm254.7}{-9.8} \\ \\ x^{\prime}=\frac{-250-254.7}{-9.8}=\frac{-504.7}{-9.8}=51.5 \end{gathered}[/tex]Note: We don't need to calculate x'' because it will be negative, and we don't have a negative time.
As the coefficient A is negative, it means that the concavity of the parabola will be facing downwards and this function will have a maximum point. So, let's use the formula below to obtain the height:
[tex]\begin{gathered} X_M=\frac{-b}{2\cdot a}=\frac{-250}{2\cdot(-4.9)}=\frac{-250}{-9.8}=25.51 \\ \\ Y_M=-\frac{b^2-4\cdot a\cdot c}{4\cdot a}=\frac{250^2-4\cdot(-4.9)\cdot122}{4\cdot(-4.9)}=\frac{64452}{-9.8}=6576.73 \end{gathered}[/tex]Answers:The rocket splashes after 51.5 seconds.
The rocket peaks at 6,576.73 meters above sea level.
