MATH SOLVE

5 months ago

Q:
# a potato is fired from a spud gun at a height of 3M and an initial velocity of 25 meters per second, write the equation of this potato projectile. How high does the potato reach and at what time does this occur?

Accepted Solution

A:

check the picture below, the height there uses feet, but nevermind that, we can simply let it be meters, like in this case.

[tex]\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{25}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{3}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-4.9t^2+25t+3[/tex]

how hight does it go?Β well, that'd be the y-coordinate of the vertex.

how long does it take?Β well, that'd be the x-coordinate of the vertex.

[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ \begin{array}{lcccl} h(t) = & -4.9t^2& +25t& +3\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{25}{2(-4.9)}~~,~~3-\cfrac{25^2}{4(-4.9)} \right)\implies \left( \cfrac{25}{9.8}~~,~~3-\cfrac{625}{-19.6} \right) \\\\\\ \left( \cfrac{25}{9.8}~~,~~3+\cfrac{625}{19.6} \right)\implies \left( \cfrac{25}{9.8}~~,~~\cfrac{3419}{98} \right)[/tex]

[tex]\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{25}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{3}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-4.9t^2+25t+3[/tex]

how hight does it go?Β well, that'd be the y-coordinate of the vertex.

how long does it take?Β well, that'd be the x-coordinate of the vertex.

[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ \begin{array}{lcccl} h(t) = & -4.9t^2& +25t& +3\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{25}{2(-4.9)}~~,~~3-\cfrac{25^2}{4(-4.9)} \right)\implies \left( \cfrac{25}{9.8}~~,~~3-\cfrac{625}{-19.6} \right) \\\\\\ \left( \cfrac{25}{9.8}~~,~~3+\cfrac{625}{19.6} \right)\implies \left( \cfrac{25}{9.8}~~,~~\cfrac{3419}{98} \right)[/tex]