Question 32164: Help!
If a soccer ball is kicked straight up from the gorund with an initial velocity of 32 feet per second, then its height above the earth in feet is given by s(t)=-16t^2 where t is time in seconds.
Graph this parabola for 0
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLES AND TRY.
YOUR FORMULA IS NOT CORRECT..IT SHOULD BE S(T)=U*T-16*T^2..WHERE U = INTIAL VELOCITY WITH WHICH THE BALL IS THROWN UP.=32 IN THIS CASE..SO S=32*T-16*T^2
MAXIMUM HEIGHT REACHED IS WHEN T=1 SEC...OR S=32-16=16 FT.
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Linear_Equations_And_Systems_Word_Problems/30576: A circus clown at the top of a 60-foot platform is playing catch with another clown on the ground.
The clown on the platform drops a ball at the same time as the one on the ground tosses a ball upward at 80 ft/sec.
For what length of time is the distance between the balls less than or equal to 10 feet? Hint: the initial velocity of a ball that is dropped is 0 ft/sec. The formula for the height S in feet above the earth at a time t seconds for an object projected into the air with an initial velocity of v ft/sec from an initial height of s0 is:
S = -16t2 + v0t + s0
1 solutions
Answer 17287 by venugopalramana(1167) About Me on 2006-03-18 11:02:08 (Show Source):
BASIS..START TIME ..0 SEC.......GROUND LEVEL =0 FT.ALL HEIGHTS MEASURED FROM GROUND LEVEL.
WE HAVE 2 BODIES B1 COMING DOWN FROM TOP TO GROUND...SAY.... U TO P.
B2 GOING UP FROM GROUND UPWARDS......SAY....P TO U....(OFCOURSE NOT CLASHING)
ATTACHMENTT:-
LET B1 COME DOWN FROM U TO R AND B2 GO UP FROM P TO Q WHEN THEY ARE AT THE 10 FEET SEPERATION ASKED FOR.THIS IS THE ATTACHMENT LET US SAY.LET THIS HAPPEN IN T1 SECS.FROM START.WE HAVE
PU=60 FEET...LET PQ=X FT.....QR=10 FT.....HENCE PR=X+10 FEET.
DISTANCE TRAVELLED IN T SECS.FROM START,MEASURED FROM GROUND LEVEL IS GIVEN BY
S=-16T^2+(V0)T+(S0).....WHERE V0 IS INITIAL VELOCITY AND S0 IS START DISTANCE FROM GROUND.
SO FOR B1 COMING DOWN WE HAVE
X+10=-16T1^2+0*T1+60=-16T1^2+60...............................I
FOR B2 GOING UP ,WE HAVE...
X=-16T1^2+80T1+0=-16T1^2+80T1..................................II
SUBSTITUTING FOR X FROM EQN.II IN EQN.I,WE GET
-16T1^2+80T1+10=-16T1^2+60
80T1=50
T1=50/80=5/8 SECS.
SO THE ATTACHMENT TAKES PLACE T 5/8 SEC. FROM START.
DETACHMENT:-
NOW B1 CONTINUES TO COME DOWN AS B2 GOES UP,CROSSING EACH OTHER,WHEN THE DISTANE BETWEEN THEM COMES DOWN FROM 10 FT. TO ZERO.AS THEY FURTHER CONTINUE THEIR TRAVEL ,THEIR DISTANCE OF SEPERATION INCREASES FROM ZERO TO 10 FT.LET US CALL THIS DETACHMENT.LET THIS HAPPEN AFTER T2 SECS.FROM START WHEN B1 REACHES SAY S AND B2 REACHES T.WE HAVE..
LET PT=Y AND HENCE PS=Y-10...SO FOR THIS PART
FOR B1 COMING DOWN WE HAVE..
Y-10=-16T2^2+0*T2+60=-16T2^2+60...........................III
FOR B2 GOING UP,WE HAVE..
Y=-16T2^2+80T2+0=-16T2^2+80T2...........................IV
SUBSTITUTING FOR Y FROM EQN.III IN EQN.IV..WE GET
-16T2^2+80T2-10=-16T2^2+60
80T2=70
T2=7/8 SECS.
SO THE DETACHMENT TAKES PLACE AT 7/8 SECS.FROM START.
SO FOR 7/8-5/8=2/8=1/4 SECS.THE 2 BALLS ARE AT A DISTANCE OF 10 FEET OR LESS.
Quadratic_Equations/30563: Solve the problem.
An object is propelled vertically upward rom the top af a112-foot building. the quadratic function s(t) = 16t^2 + 176t + 112 models the ball's height above the ground, s(t), in feet, t seconds after it was thrown. How many seconds does it take until the object finally hits the ground. Round to the nearest tenth of a second if necessary.
Please help me. I am just not getting this. Thank you.
1 solutions
Answer 17241 by venugopalramana(1167) About Me on 2006-03-17 20:47:03 (Show Source):
30563An object is propelled vertically upward rom the top af a112-foot building. the quadratic function s(t) = 16t^2 + 176t + 112 models the ball's height above the ground, s(t), in feet, t seconds after it was thrown. How many seconds does it take until the object finally hits the ground. Round to the nearest tenth of a second if necessary.
THERE IS NOTING AMISS IN YOU NOT GETTING THE SOLUTION IF YOU ARE NOT TAUGHT CALCULUS...WE CAN CIRCUMVENT THAT BY USING SOME CIRCUTOUS PROCEDURE.BUT YOU TELL ME FIRST WHETHER YOU KNOW DIFFERENTIATION...SEE BELOW.
FURTHER I FEEL THE FORMULA IS BETTER REPRESENTED BY S= -16T^2+176T+112.(YOU WILL FIND THAT THIS ELIMINATES NEGATIVE TIMINGS OBTAINED BELOW.)
WE HAVE TO CONSIDER 2 MOTIONS HERE.
1.UPWARD...THE OBJECT GOES UP TILL ITS VELOCITY BECOMES ZERO.LET IT GO A DISTANCE H DURING THIS PART.
2.DOWNWARD.....THE OBJECT FALLS ON TO THE GROUND.DURING THIS PHASE IT HAS TO TRAVEL H..THE SAME DISTANCE IT HAS GONE UP PLUS 112 FT THE HEIGHT OF THE BUILDING...SO DISTANCE TRAVELLED DURING DESCENT =H+112
3.HENCE TOTAL DISTANCE TRAVELLED IS H+H+112=2H+112
s(t) = 16t^2 + 176t + 112
VELOCITY =DS/DT =32T+176
WHEN THE OBJECT REACHES ITS TOP POSITION ,ITS VELOCITY BECOMES ZERO AND IT STARTS TO FALL DOWN
HENCE TIME TAKEN TO REACH PEAK IS GIVEN BY VELOCITY = 0 = 32T+176
T=176/32= - 5.5 …MINUS IS BECAUSE ,THE GIVEN FORMULA DOES NOT DISTINGUISH FOR UPWARD/DOWNWARD MOTION.WE CAN TAKE ABSOLUTE VALUE FOR OUR PURPOSE. IT TAKES SAME TIME FOR DECENT UPTO TOP OF BUILDING.HENCE
TIME FOR DOWNWARD MOVEMENT UPTO TOP OF BUILDING = 5.5
DISTANCE FROM TOP OF BUILDING TO GROUND = 112.SO TIME FOR THIS TRAVEL IS GIVEN BY....S=112 = 16T^2+176T+112
16T(T+11)=0..HENCE T+11=0....OR T= -11
SO TOTAL TIME TAKEN IS = 5.5+5.5+11 = 22...
NOW ON TO SOLUTION WITHOUT CALCULUS...THAT IS SOLVING FOR TIME TAKEN TO REACH THE PEAK FROM BUILDING TOP AND COMING BACK TO TOP OF BUILDING WITHOUT DIFFERENTIATION TO GET AT VELOCITY.
YOU CAN TREAT THE GIVEN EQN.AS A QUADRATIC WITH A PEAK REPRESENTING THE TOP POSITION OF OBJECT AS EXPLAINED ABOVE.THIS HELPS TO GET THE TIME TAKEN TO REACH THE PEAK AS A PHYSICAL INTERPRETATION OF THE FORMULA/PHENOMENA.
S=16T^2+176T+112=16{T^2+11T+7)=16{(T^2+2T(11/2)+(11/2)^2)-(11/2)^2+7)}
=16{(T+5.5)^2-93/4}....SO THE PEAK OCCURS AT
T+5.5=0..OR...T=-5.5...THE NEGATIVE SIGN WAS ALREADY EXPLAINED ABOVE...
THUS YOU CAN SHOW T=5.5 FOR ASCENT +5.5 FOR DESCENT.FURTHER TRAVEL BY S=112 FROM TOP OF BUILDING CAN BE DONE BY SAME WAY AS ABOVE AS IT DOES NOT INVOLVE ANY CALCULUS.
Travel_Word_Problems/30292: Please answer this question for me: A missle is fired with an initial velocity of 1600 feet per second. When will it be 40,000 feet above the starting point?
The answer I got was: 25 seconds because I did 40000
------ = 25
I think its wrong though 1600
1 solutions
Answer 16973 by venugopalramana(1167) About Me on 2006-03-14 23:02:42 (Show Source):
SEE THE FOLLOWING AND DO
IN FPS SYSTEM ....G=32 FT/SEC^2..SO USE -32 INSTEAD OF -9.8 IN THE FOLLOWING
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PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point?
Than answer I got was : 40 second because I divided... I think its wrong thought
AS THE ROCKET GOES UP AGAINST EARTHS GRAVITATIONAL FORCE , ITS VELOCITY REDUCES,GIVEN BY ACCELERATION DUE TO GRAVITY =-9.8 M/SEC^2..THE FORMULA TO BE USED IS
S = UT + (G/2)T^2....WHERE
U = INIYIAL VELOCITY = 800 M/SEC
T = TIME ELAPSED IN SECS.
G=ACCELERATION DUE TO GRAVITY = - 9.8 M/SEC^2 FOR UPWARD TRAVEL...SO..WE GET
32000 = 800T -(9.8/2)T^2=800T-4.9 T^2
4.9T^2-800T+32000 =0
t = (800 +- sqrt( 800^2-4*4.9*32000 ))/(2*4.9)
T=70.1 WHILE GOING UP AND
T=93.2 WHILE FALLING DOWN.
Travel_Word_Problems/30301: PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point?
Than answer I got was : 40 second because I divided... I think its wrong thought
1 solutions
Answer 16965 by venugopalramana(1167) About Me on 2006-03-14 22:06:23 (Show Source):
PLEASE HELP ME A rocket is fired with an initial velocity of 800 meters per second. When will it be 32000 meters above the starting point?
Than answer I got was : 40 second because I divided... I think its wrong thought
AS THE ROCKET GOES UP AGAINST EARTHS GRAVITATIONAL FORCE , ITS VELOCITY REDUCES,GIVEN BY ACCELERATION DUE TO GRAVITY =-9.8 M/SEC^2..THE FORMULA TO BE USED IS
S = UT + (G/2)T^2....WHERE
U = INIYIAL VELOCITY = 800 M/SEC
T = TIME ELAPSED IN SECS.
G=ACCELERATION DUE TO GRAVITY = - 9.8 M/SEC^2 FOR UPWARD TRAVEL...SO..WE GET
32000 = 800T -(9.8/2)T^2=800T-4.9 T^2
4.9T^2-800T+32000 =0
t = (800 +- sqrt( 800^2-4*4.9*32000 ))/(2*4.9)
T=70.1 WHILE GOING UP AND
T=93.2 WHILE FALLING DOWN.
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