SOLUTION: PROBLEM: HEIGHT OF A PRJECTILE: A PROJECTILE IS LAUNCHED FROM GROUND LEVEL WITH AN INITIAL VELOCITY OF v0 FEET PER SECOND. NEGLECTING AIR RESISTANCE, ITS HEIGHT IN FEET t SECONDS

Algebra ->  Triangles -> SOLUTION: PROBLEM: HEIGHT OF A PRJECTILE: A PROJECTILE IS LAUNCHED FROM GROUND LEVEL WITH AN INITIAL VELOCITY OF v0 FEET PER SECOND. NEGLECTING AIR RESISTANCE, ITS HEIGHT IN FEET t SECONDS       Log On


   

Question 143954This question is from textbook College Algebra
: PROBLEM: HEIGHT OF A PRJECTILE:
A PROJECTILE IS LAUNCHED FROM GROUND LEVEL WITH AN INITIAL VELOCITY OF v0 FEET PER SECOND. NEGLECTING AIR RESISTANCE, ITS HEIGHT IN FEET t SECONDS AFTER LAUCH IS GIVEN BY s = -16^2+ v0t
::: IN NUMBER 25, FIND THE TIME (s) THAT THE PROJECTILE WILL (a)REACH A HEIGHT OF 80ft AND (b) RETURN TO THE GROUND FOR THE GIVEN VALUE OF v0. ROUND ANSWERS TO THE NEAREST HUNDREDTH IF NECESSARY.
** ON THIS PROBLEM I DO NOT KNOW EVEN WHERE TO BEGIN..? ALL OF THE SIGN AND WORDING IS VERY DIFFICULT FOR ME TO COMPREHEND.. PLEASE HELP??.. :) This question is from textbook College Algebra

Found 2 solutions by stanbon, vleith:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
s(t) = -16t^2+ vot
Comment: s(t) is the height of the projectile after t seconds.
vo=96 ft/sec is the velocity of the projectile when you let go of it.
::: IN NUMBER 25, FIND THE TIME (t) THAT THE PROJECTILE WILL
(a)REACH A HEIGHT OF 80ft :
80 = -16t^2 + 96t
-16t^2+96t-80= 0
-16(t^2-6t+5)=0
(t-1)(t-5)=0
t = 1 or t=5
The projectile reaches a height of 80 ft. after 1 second (on the way up)
and at 5 seconds (on the way down).
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(b) RETURN TO THE GROUND FOR THE GIVEN VALUE OF v0=96 ft/sec.
ROUND ANSWERS TO THE NEAREST HUNDREDTH IF NECESSARY.
-16t^2+96t=0
-16t(t-6)=0
t = 0 or t=6
It's on the ground at time=0 and again at time = 6 seconds.
===============
Cheers,
Stan H.
=================
Cheers,
Stan H.

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
You are given a equation that relates distance (s) to both time (t) and initial velocity (V0)
s+=+-16t%5E2+%2B+v%5B0%5Dt
The first question asks you to find the equation for time when s = 80 for a given v0.
80+=+-16t%5E2+%2B+v%5B0%5Dt
0+=+-16t%5E2+%2B+v%5B0%5Dt+-+80
Now use the quadratic equation with a=-16, b=vo and c = -80
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Note that you won't get 'an answer' that is a number since the value v0 is not known. If it turns out that there is only one time when s=80, then you know that 80 feet up is as high as the projectile goes. If there are two times, then one is on the way up and the other is on the way back down. You won't know yet since you have not been given a value for V0. But you can find what v0 will result in s=80 being the highest point (just find when sqrt%28+b%5E2-4%2Aa%2Ac+%29 is 0
Either way, I suggest you move from where the projectile was fired from, because it is going to return there :) When will it return there? Well that is problem 2.
When will it hit the ground?
Use the same logic as in problem 1, except instead of 80 feet, use 0. There will be two answers for 0. one will be at time 0, the other is the one they want.