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: the t intercept b) what is the maximum height the rocket reaches? to answer this question, wed find: select an answer c) if the rocket will splash down in down? to answer this question, wed find: question help: video written the t intercept the h intercept the t coordinate of the vertex the h coordinate of the vertex submit question

Explanation:

Part a)

To find the launch height, we need the height when \( t = 0 \) (launch time). The \( h \)-intercept (when \( t = 0 \)) of the function \( h(t)=- 4.9t^{2}+1200t + 440 \) gives this value. However, the original selection of "The \( t \)-intercept" is incorrect. The correct choice should be "The \( h \)-intercept" (since \( t = 0 \) gives the initial height). But if we follow the problem's initial (wrong) dropdown and just state the intended (even if wrong in the problem's initial) - but actually, to correct: At \( t = 0 \), \( h(0)=-4.9(0)^{2}+1200(0)+440 = 440 \). So the launch height is found by the \( h \)-intercept (when \( t = 0 \)). But the problem's initial dropdown had "The \( t \)-intercept" which is wrong. But if we answer based on the problem's question (what to find for launch height), the correct is the \( h \)-intercept (when \( t = 0 \)), but the problem's initial dropdown was wrong. However, if we take the problem's context, maybe a typo. But strictly, for launch height (t=0), it's the h-intercept (value of h when t=0).

The function \( h(t)=-4.9t^{2}+1200t + 440 \) is a quadratic function in the form \( y = ax^{2}+bx + c \) with \( a=-4.9<0 \), so it opens downward. The vertex of a parabola \( y = ax^{2}+bx + c \) has its maximum (since \( a<0 \)) at the vertex. The \( h \)-coordinate of the vertex gives the maximum height. The formula for the \( y \)-coordinate (here \( h \)-coordinate) of the vertex is \( h=-\frac{b^{2}-4ac}{4a} \) or we can first find the \( t \)-coordinate of the vertex \( t =-\frac{b}{2a} \) and then plug into \( h(t) \). But to find the maximum height, we need the \( h \)-coordinate of the vertex.

Splash down occurs when the height \( h(t)=0 \) (since it's at sea level). So we need to find the value of \( t \) when \( h(t) = 0 \), which is the \( t \)-intercept (the value of \( t \) where the graph of \( h(t) \) crosses the \( t \)-axis, i.e., \( h(t)=0 \)).

Answer:

The correct option (to find launch height) should be "The \( h \)-intercept" (since at \( t = 0 \), \( h(0) \) is the initial height). But the problem's initial dropdown had "The \( t \)-intercept" which is incorrect. If we have to choose from the options (even if the initial was wrong), but the options for part a) dropdown was "The \( t \)-intercept" (wrong), but the correct mathematical approach is \( h(0) \), so the \( h \)-intercept. But since the problem's part a) dropdown is given as "The \( t \)-intercept" (incorrect), but maybe a mistake.

Part b)