Respuesta :
[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{39.2}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{\textit{from the ground, so }0}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-4.9t^2+39.2t+0\implies h(t)=-4.9t^2+39.2t[/tex]
The equation that can be used to model the height of the rocket after t seconds is h = 39.2t -4.9t².
What are the equation of motion?
There are three equations of motion, v = u +at, v² = u² +2as, and s = ut+(1/2) at².
The rocket is launched from the ground ,
Initial Height = 0
Initial Velocity = 39.2 m/sec
The equation used to model the height of the rocket is
s = ut+(1/2) at²
h = 39.2 t + (1/2) at²
The acceleration is of the gravity g, but in opposite direction of the motion so a = -9.8 m/s²
h = 39.2t -4.9t²
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