Question 709069
{{{s= 144+128t-16t^2}}}
a) what is the height of the rocket 5 seconds after it is launched?
This is just another way of asking: "What is s when t = 5?" For this we just substitute 5 for t and solve for s:
{{{s= 144+128(5)-16(5)^2}}}
Simplifying...
{{{s= 144+128(5)-16(25)}}}
{{{s= 144+640-400}}}
{{{s = 384}}}
So after 5 seconds the height, s, is 384 feet.<br>
B) after how many second does the rocket splash in the water?
Since the water represents a height of 0, this is just another way of asking: "At what time(s) will the height, s, be zero?" For this we will replace the s with 0 and solve for t:
{{{(0)= 144+128t-16t^2}}}
This is a quadratic equation (which often has two solutions). So I will start by putting it in standard {{{ax^2+bx+c}}} form:
{{{0= -16t^2+128t+144}}}
Since one side of the equation is already zero we can proceed to the next step. We factor (or use the Quadratic Formula). The GCF is 16. So we can factor out 16 or -16. Since factoring out -16 will leave the {{{t^2}}} with a positive coefficient and since the positive coefficient makes the rest of the factoring easier I'm going to factor out -16:
{{{0= -16(t^2-8t-9)}}}
The second factor factors fairly easily (since we thought to factor out the "-" from in front of {{{t^2}}}):
{{{0= -16(t-9)(t+1)}}}
From the Zero Product Property we know that one (or more) of the factors must be zero. So:
-16 = 0 or t-9 = 0 or t+1 = 0
The first equation is a false statement so it has no solutions. The other two, however, have solutions:
t = 9 or t = -1<br>
Since t represents time we will reject the negative answer. (-1 as t means "one second before we launched the rocket..." which makes no sense.) So there is only one realistic answer to the question. The rocket will hit the water 9 seconds after the launch.