If an object is dropped from a height of 144 feet, the function h(t)= -16t^2+144 gives the height of the object after t seconds. When will the object hit the ground? A. 9 s B. 6 s C. 1.5 s D. 3 s
someone please help i'm confused
When it hits the ground, what is its height?
is the answer d
Not what I asked.
0 i think
Then let zero equal h(t), and solve for t.
so the answer is d
because -16*9+144=0
You were correct before. \[h(t)= -16t^2+144=0 \implies 16t^2=144 \implies t^2=9 \implies t=3\]
thanks
Nice work. Do math every day.
thats great
can u walk me through another one please?
Sure.
hold on
A catapult launches a boulder with an upward velocity of 112 ft/s. The height of the boulder, h, in feet after t seconds is given by the function . How long does it take the boulder to reach its maximum height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary. h=-16t^2+112t+30 A. 7 s; 30 ft B. 3.5 s; 366 ft C. 3.5 s; 618 ft D. 3.5 s; 226 ft
OK, this one is a little different from the first. What kind of function is h(t)?
i'm confused
Look at the function. What kind of function is f(x)=ax^2+bx+c?
idk
Quadratic function. You are obviously learning about them, since both of your problems have had them.
Does the graph of h(t) open up or down?
ok
down
Good. Now, since h(t) is the height at a given time, we want the point on that graph where the height is a maximum. Do you understand that from the problem?
no
Note that the two parts of the question are "When is the boulder at maximum height?" and "What is the maximum height?" (paraphrased here....). So, if we think of the graph, it is the vertex of the parabola. Getting it now?
yes
The t value of the max height can be found by using t=(-b/2a) from the quadratic function you have. After substitution, we get t=-112/-32=7/2. Plug 7/2 into the function to get the maximum height. I'll let you do that part now....
I get about 226 feet. On your list of possible answers, that is answer d.
ok thanks
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