Question

The height h of an arrow in feet is modeled by h(t) = -16t2 + 63t + 4, where t is the time in seconds since the arrow was shot. How long is the arrow in the air?

Answer 1

so if I were you, I will find dh/dt = maximum of the resulting curve. Such maximum occurs ath the highest point of the given expression, where the velocity is zero. Then I proceed by solving dh(t)/dt = 0. This probably given me the value of t at this point, right? Then I assume that the time taken to arrive at this point corresponds to the value obtained at the maximum!

Answer 2

so do I first divide 63/-16?

Answer 3

No its a calculus question. U find the derivative of H(t) with respect to t. This gives you the velocity. Did you guys study calculus?

Answer 4

Not really. This is my first application question I have done for this section.

Answer 5

So where would i start solving this. Because I am really confused

Answer 6

So you never heard about the derivative or what?

Answer 7

no

Answer 8

ok lets see u might need to solve the h(t) = 0 and probably divide by 2 or something. This could give u the time it takes. Hang on a sec lemme read the question again.

Answer 9

how would you solve that? /.\ h(t)=0/2?

Answer 10

notice that the leading coefficient of the parabola equation is negative, thus is opening DOWNWARD, so

Answer 11

Yes because the arrow would be falling

Answer 12

those would be the x-intercepts, so set y = 0, and solve for "x", like for any quadratic equaation \(\bf h(t) = -16t2 + 63t + 4\implies 0 = -16t2 + 63t + 4\)