Question

A rocket is launched from the top of a 55-foot cliff with an initial velocity of 138 ft/s. a. Substitute the values into the vertical motion formula h = –16t2 + vt + c. Let h = 0. b. Use the quadratic formula to find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.

Answer 1

check the picture below.


`\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{138}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{55}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at`


`\bf ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+138}t\stackrel{\stackrel{c}{\downarrow }}{+55} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t=\cfrac{-138\pm√(138^2-4(-16)(55))}{2(-16)} \\\\\\ t=\cfrac{-138\pm√(138^2+3520)}{2(-16)}\implies t=\cfrac{-138\pm√(22546)}{-32} \\\\\\ t\approx \begin{cases} -0.38166193265635\\ \boxed{9.00666193265635} \end{cases}`