SOLUTION: The height of a projectile fired upward from the ground with a velocity of 128 feet per second is given by the formula: h=-16t squared + 128t. a. When will the projectile be 256

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Question 1156441: The height of a projectile fired upward from the ground with a velocity of 128 feet per second is given by the formula: h=-16t squared + 128t.
a. When will the projectile be 256 feet above the ground?
b. Will the projectile ever be 300 feet above the ground? Explain.
c. When will the projectile be 240 feet above the ground?
d. In how many seconds will the projectile hit the ground?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

The height of a projectile fired upward from the ground with a velocity of 128 feet per second is given by the formula:
h=-16t%5E2+%2B+128t
a. When will the projectile be 256 feet above the ground?
h=256
256=-16t%5E2+%2B+128t
16t%5E2-128t%2B256=0+....both sides divide by 16
t%5E2-8t%2B16=0+......factor
t%5E2-4t-4t%2B16=0+
%28t%5E2-4t%29-%284t-16%29=0+
t%28t-4%29-4%28t-4%29=0+
%28t-4%29%28t-4%29=0+
=>t=4
the projectile will be 256 feet above the ground in 4 seconds


b. Will the projectile ever be 300 feet above the ground? Explain.
h=300
300=-16t%5E2+%2B+128t
16t%5E2-128t%2B300=0+
use quadratic formula:
t=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a...in your case a=16,b=-128, and c=300
t=%28-%28-128%29%2B-sqrt%28%28-128%29%5E2-4%2A16%2A300%29%29%2F%282%2A16%29
t=%28128%2B-sqrt%2816384-19200%29%29%2F32
t=%28128%2B-sqrt%28-2816%29%29%2F32=> since determinant is negative number, we have imaginary roots
=> the projectile will never be 300 feet above the ground

c. When will the projectile be 240 feet above the ground?
h=240
240=-16t%5E2+%2B+128t
16t%5E2-128t%2B240=0+
16%28t%5E2-8t%2B15%29=0+
use quadratic formula:
t=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a...in your case a=1,b=-8, and c=15
t=%28-%28-8%29%2B-sqrt%28%28-8%29%5E2-4%2A1%2A15%29%29%2F%282%2A1%29
t=%288%2B-sqrt%2864-60%29%29%2F2
t=%288%2B-sqrt%284%29%29%2F2
t=%288%2B-2%29%2F2
solutions:
t=%288%2B2%29%2F2=>t=5
t=%288-2%29%2F2=>t=3
the projectile will be 240 feet above the ground in 3 and again in 5
d. In how many seconds will the projectile hit the ground?
h=0
0=-16t%5E2+%2B+128t
16t%5E2+-128t=0
16t%28t%5E2+-8%29=0
if 16t=0=>t=0=> not solution
if t%5E2+-8=0=> t%5E2+=8=> t+=sqrt%288%29=> t+=2.83
the projectile hit the ground in 2.83 seconds


Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The last part d) in the post by @MathLover1 should be corrected. 


-16*t^2 + 128t = 0


16t*(t-8) = 0


t = 0  is not the solution.


t = 8   is the ANSWER.

Solved (correctly).

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In this site,  there is a bunch of lessons on a projectile thrown/shot/launched vertically up
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

Consider these lessons as your textbook,  handbook,  tutorials and  (free of charge)  home teacher.
Read them attentively and learn how to solve this type of problems once and for all.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.