SOLUTION: I am trying to help my sister with her homework and I am stumped. Please help. here is the question. (this is a take home test she has failed once already and was told to re-do it.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am trying to help my sister with her homework and I am stumped. Please help. here is the question. (this is a take home test she has failed once already and was told to re-do it.      Log On


   

Question 660330: I am trying to help my sister with her homework and I am stumped. Please help. here is the question. (this is a take home test she has failed once already and was told to re-do it. She skipped all of these because she doesnt understand them and her teacher wont help her.)
Three objects are launched from the top of a 160ft platform. The first object is launched downward at 48ft/sec. The second object is dropped. The third object is launched upward at 96ft/sec. Write a height model for the first, second and third object. And determine the length of time the third object was in the air.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In words:
You need an equation that has 3 parts
to describe the height above ground:
h = ( height change due to throwing ) - ( height change due to gravity )
+ ( the initial height above ground )
-----------------------
Actual formula looks like:
+h+=+v%5B0%5D%2At+-+16t%5E2+%2B+h%5B0%5D+
------------------------
(a)
The first object is launched downward at 48ft/sec.
That means the +v%5B0%5D%2At+ term is minus, the
same as gravity term
+h+=+-v%5B0%5D%2At+-+16t%5E2+%2B+h%5B0%5D+
+h%5B0%5D+=+160+ ft
+v%5B0%5D+=+48+ ft/sec downward
+h+=+-48t+-+16t%5E2+%2B+160+
------------------------------
(b)
The second object is dropped.
That means the +v%5B0%5D%2At+ term is zero ( no throwing )
+h+=+-16t%5E2+%2B+160+
------------------------------
(c)
The third object is launched upward at 96ft/sec.
The +v%5B0%5D%2At+ term is positive
+h+=+96t+-+16t%5E2+%2B+160+
-------------------------------
determine the length of time the third object was in the air.
That means the object is trhown up, reaches a peak, and
comes back to the ground
Note that when +t+=+0+, the equation becomes
+h+=+160+ which means object hasn't been thrown yet
You want to find +t+ when +h+=+0+ ( back to ground )
+0+=+-16t%5E2+%2B+96t+%2B+160+
Divide both sides by +16+
+0+=+-t%5E2+%2B+6t+%2B+10+
Use quadratic formula
+t+=+%28+-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+-1+
+b+=+6+
+c+=+10+
+t+=+%28+-6+%2B-+sqrt%28+36+-+4%2A%28-1%29%2A10+%29%29%2F%282%2A%28-1%29%29+
+t+=+%28+-6+%2B-+sqrt%28+36+%2B+40+%29%29%2F%28-2%29+
+t+=+%28+-6+%2B-+sqrt%28+76+%29%29%2F%28-2%29+
+t+=+%28+-6+-+sqrt%28+76+%29%29+%2F+%28-2%29+
This gives the time in the air
Here's the plot of the equation +h+=+-16t%5E2+%2B+96t+%2B+160+
You can see the height of +160+ ft where object is thrown
up and the time of 7+ sec when it hits ground
+graph%28+400%2C+400%2C+-2%2C+10%2C+-20%2C+320%2C+-16x%5E2+%2B+96x+%2B+160+%29+