SOLUTION: A man is standing on a rock and shoots a firework into the air. The path of the firework can be modelled by the equation h(x)=-5x^2 +7x +3, where h(x) is height in meters and x is

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: A man is standing on a rock and shoots a firework into the air. The path of the firework can be modelled by the equation h(x)=-5x^2 +7x +3, where h(x) is height in meters and x is       Log On


   

Question 1057359: A man is standing on a rock and shoots a firework into the air. The path of the firework can be modelled by the equation h(x)=-5x^2 +7x +3, where h(x) is height in meters and x is time in seconds.
Must find height rocket was released from the ground
Max height rocket reaches
How many seconds it reaches its max height
After how many seconds does it hit the ground
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.
See the lessons
    - Problem on a projectile moving vertically up and down
    - Problem on a toy rocket launched vertically up from a tall platform
in this site.

Very similar problems were solved there. They are your samples/prototypes.

Read these lessons and then solve your problem by substituting your data.

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".


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+h%28x%29+=+-5x%5E2+%2B+7x+%2B+3+
The rocket is launched at +x+=+0+
so, +h%280%29+ gives you the height the
rocket was launched from
+h%280%29+=+-5%2A0%5E2+%2B+7%2A0+%2B+3+
+h%280%29+=+3+
Height at launch time was 3 m
------------------------------
The x-value of the vertex +h%5Bmax%5D+ is
given by the formula:
+x%5Bmax%5D+=+-b%2F%282a%29+, where
+a+=+-5+
+b+=+7+
-----------------
+x%5Bmax%5D+=+-7%2F%282%2A%28-5%29%29+
+x%5Bmax%5D+=+7%2F10+
Plug this value back into equation to
find +h%5Bmax%5D+
+h%5Bmax%5D+=+-5%2A%287%2F10%29%5E2+%2B+7%2A%287%2F10%29+%2B+3+
+h%5Bmax%5D+=+-5%2A%2849%2F100%29+%2B+49%2F10+%2B+60%2F20+
+h%5Bmax%5D+=+-49%2F20+%2B+98%2F20+%2B+60%2F20+
+h%5Bmax%5D+=+109%2F20+
+h%5Bmax%5D+=+5.45+ m
The max height is 5.45 m
This is only 2.45 m above launch height.
Something seems wrong. I think the +7x+
term should be a lot bigger.
-------------------------------
The time when the rocket reaches max height
is the 7/10 sec
-------------------
The rocket hits the ground when +h%28x%29+=+0+
( zero height )
+-5x%5E2+%2B+7x+%2B+3+=+0+
Use the quadratic formula
+x+=+%28+-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+-5+
+b+=+7+
+c+=+3+
+x+=+%28+-7+%2B-+sqrt%28+7%5E2-4%2A%28-5%29%2A3+%29%29%2F%282%2A%28-5%29%29+
+x+=+%28+-7+%2B-+sqrt%28+49+%2B+60+%29%29%2F%28-10%29+
+x+=+%28+-7+%2B-sqrt%28+109+%29%29+%2F+%28-10%29+
+x+=+%28+-7+-+10.44+%29+%2F+%28-10%29+
+x+=+17.44+%2F+10+
+x+=+1.744+
In 1.744 sec the rocket hits the ground
--------------------------------------
Here's the plot of the equation:
+graph%28+400%2C+400%2C+-1%2C+3%2C+-1%2C+7%2C+-5x%5E2+%2B+7x+%2B+3+%29+
This doesn't look like the flight of a "rocket". Check
your data and my math ( I think my math is OK )