Question

A rocket carrying fireworks is launched form a hill 80 above a lake. The rocket will fall into the lake after exploding at its maximum height. The rocket's height above the surface of the lake is given by: h left parenthesis t right parenthesis equals space minus 16 left parenthesis t minus 2 right parenthesis squared plus 144

Answer 1

Answer: After 5 s

Step-by-step explanation:

We are told the equation that models the height of the rocket is:

$h=-16(t-2)^{2}+144$ (1)

Where $h$ is the height and $t$ is the time.

This equation can be rewritten as follows:

$h=-16(t^{2}+4 t+4)+144$ (2)

$h=16t^{2}+64 t+80$ (3)

So, we have to find the time at which the rocket hits the lake, this means when $h=0$:

$0=-16t^{2}+64 t+80$ (4)

Dividing both sides of the equation by 16:

$0=t^{2}+4 t+5$ (5)

Multiplying both sides of the equation by -1:

$0=t^{2}-4 t-5$ (6)

This is a quadratic equation of the form $0=at^{2}+bt+c$, and we have to use the quadratic formula if we want to find  $t$:  

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (7)

Where $a=1$, $b=-4$, $c=-5$  

Substituting the known values and choosing the positive result of the equation:  

$t=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-5)}}{2(1)}$  

$t=5 s$  (8)

Hence, the rocket hits the lake 5 seconds after its launch.