SOLUTION: I do not understand these problems... An arrow is shot upward at an initial velocity of 40 meters per second. Use the function: h=40t-5t^2 to answer the following questions:

Question 1159348: I do not understand these problems...
An arrow is shot upward at an initial velocity of 40 meters per second. Use the function: h=40t-5t^2 to answer the following questions:
1).what is the height of the arrow in meters after 5 seconds? Solve algebraically. Show Work.
2). After how many seconds will the arrow be 30 meters high? Use the quadratic formula. Show work.
3). when will the arrow be back on the ground? solve algebraically using factoring. Show work.
Thank you for your help
Found 3 solutions by Boreal, MathLover1, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Draw a parabola to show the path of the arrow.
The equation is height=-5t^2+40t, the -5 is acceleration of gravity (-4.9 more accurately) and the 40 is the speed of the arrow in m/sec. The units are (m/sec^2)(sec^2) or meters and (m/sec)*sec or meters.
t is the number of seconds
h(5)=-5(25)+40(5)=75 meters high
the arrow is 30 meters high twice, going up and coming down
30=-5t^2+40t
rewrite to make t^2 term positive. It's easier that way.
5t^2-40t+30=0
divide by 5 to simplify it
t^2-8t+6=0
quadratic formula
t=(1/2)(8+/8+/-sqrt(64-24) or (1/2)* 8 +/- sqrt (40) which is 2 sqrt(10)
t=4+/- sqrt (10) multiplying it by the 1/2
This is at t=7.16 seconds and 0.84 seconds. You can make those to more decimal places if you wish and show that they will both give 30 m when put into the equation.
it hits the ground when H(t)=0
-5t^2+40t=0
5t^2=40t
t^2=8t
t=8 seconds
If you substitute 8 into the equation, it equals 0.

Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
given the function:

1).what is the height of the arrow in meters after seconds?
........if seconds

meters high

2). After how many seconds will the arrow be meters high?
........if

......simplify, divide by
.......use quadratic formula

.........simplify

-> solutions:
-> seconds
-> seconds

so, it will ritch height of meters in seconds and second time (on its way back) seconds

3). when will the arrow be back on the ground?
the arrow be back on the ground when

->->when ...this is start position
->->when -> it will take seconds for the arrow to be back on the ground

Answer by ikleyn(52779)   (Show Source): You can put this solution on YOUR website!
.

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.

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