SOLUTION: Help please! I'm so confused by this problem. Can you explain it step by step? if an object is thrown vertically upward with an initial velocity of v, from an original position

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Help please! I'm so confused by this problem. Can you explain it step by step? if an object is thrown vertically upward with an initial velocity of v, from an original position       Log On


   

Question 763698: Help please! I'm so confused by this problem. Can you explain it step by step?
if an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h=-16t^2+vt+s
(where h and s are in ft, t is in seconds and v is in ft/sec)
a ball is thrown upward with an initial velocity of 96 ft/sec from the top of a 100 ft bridge. Determine the time that it takes for the ball to get to a height of 200 ft.
Round answers to 2 decimals.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+h%28t%29+=+-16t%5E2%2Bvt%2Bs+
notice I replaced +h+ with +h%28t%29+
It's just a matter of preference.
---------------------------
Think of the 1st term as how the force
of gravity is affecting the height.
The other terms have nothing to do with gravity.
----------------------------
The 2nd term has to do with how fast you
release the ball, and whether you throw it
up or down (+) is for up, (-) is for down.
----------------------------------
The 3rd term is only the height above ground
from which the ball is released.
---------------------------
It's important to see that at +t+=+0+, just
as the ball is released,
+h%280%29+=+-16%2A0%5E2%2Bv%2A0%2Bs+
+h%280%29+=+s+, the height from which ball is released,
and you are given that +s+=+100+
-------------------------------
You are asked to find +t+ when +h%28t%29+=+200+ ft
Note that the positive +96+ ft/sec has to fight
against the negative +16+ ft/sec2 which is acceleration
due to gravity.
------------------
+200+=+-16t%5E2+%2B+96t+%2B+100+
+-16t%5E2+%2B+96t+-+100+=+0+
+-4t%5E2+%2B+24t+-+25+=+0+
use quadratic formula
+t+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+-4+
+b+=+24+
+c+=+-25+
+t+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+t+=+%28-24+%2B-+sqrt%28+24%5E2+-+4%2A%28-4%29%2A%28-25%29+%29%29+%2F+%282%2A%28-4%29%29+
+t+=+%28-24+%2B-+sqrt%28+576+-+400+%29%29+%2F+%28-8%29+%29+
+t+=+%28-24+%2B-+sqrt%28+176+%29%29+%2F+%28-8%29+%29+
+t+=+%28-24+%2B-+13.2665+%29+%2F+%28-8%29+%29+
There are 2 valid solutions. There is a height when
the ball is on the way up to it's peak, and also a
height when the ball is on the way down from it's peak.
+t+=+%28+-24+%2B+13.2665+%29+%2F+%28-8%29+
+t+=+1.3417+ sec
and, also
+t+=+%28+-24+-+13.2665+%29+%2F+%28-8%29+
+t+=+-37.2665+%2F+%28-8%29+
+t+=+4.6583+ sec
( note that you should never end up with a negative time )
Here's a plot:
+graph%28+400%2C+400%2C+-2%2C+20%2C+-20%2C+250%2C+-16x%5E2+%2B+96x+%2B+100+%29+
At t = 1.34 sec the height is 200 ft and
At t = 4.66 sec the height is 200 ft
You can see this on plot
Hope this helps