Question

this question is not about solving the stated problem, but about understanding it. a rocket is launched, and its height above sea level t seconds after launch is given by the equation $h(t) = -4.9t^2 + 1200t + 440$. a) from what height was the rocket launched? to answer this question, wed find: select an answer b) what is the maximum height the to answer this question, wed find: c) if the rocket will splash down in down? to answer this question, wed find: question help: video written example select an answer the t intercept the h intercept the t coordinate of the vertex the h coordinate of the vertex

Explanation:

Part (a)

The height function is \( h(t) = - 4.9t^{2}+1200t + 440 \). The rocket is launched at \( t = 0 \) seconds. To find the launch height, we substitute \( t = 0 \) into the function. When \( t = 0 \), the term with \( t^{2} \) and the term with \( t \) will be zero, and we are left with the constant term. The \( h \)-intercept (the value of \( h(t) \) when \( t = 0 \)) gives the initial height. So we need to find the \( h \)-intercept.

The height function \( h(t)=-4.9t^{2}+1200t + 440 \) is a quadratic function in the form \( y = ax^{2}+bx + c \) where \( a=- 4.9\), \( b = 1200 \) and \( c = 440 \). Since \( a<0 \), the parabola opens downwards, and the vertex of the parabola represents the maximum point. The \( h \)-coordinate of the vertex gives the maximum value of the function (in this case, the maximum height). The formula for the \( h \)-coordinate of the vertex of a quadratic \( y=ax^{2}+bx + c \) is \( h=-\frac{b^{2}-4ac}{4a} \) or we can also think of it as evaluating the function at the \( t \)-coordinate of the vertex (\( t=-\frac{b}{2a} \)). But the value we want is the maximum height, which is the \( h \)-coordinate of the vertex.

The rocket splashes down when its height \( h(t)=0 \). We need to find the value of \( t \) (time) when \( h(t) = 0 \). The \( t \)-intercepts of the function \( h(t) \) are the values of \( t \) for which \( h(t)=0 \). Since the rocket is launched at \( t = 0 \) and then comes back down, we are interested in the positive \( t \)-intercept (the non - zero \( t \) value where \( h(t)=0 \)) which gives the time when the rocket splashes down. So we need to find the \( t \)-intercept.

Answer:

The h intercept

Part (b)