Question 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
Found 2 solutions by Fermat, venugopalramana:
Answer by Fermat(136)   (Show Source):
You can put this solution on YOUR website! There are two eqns of motion. One for the ball falling down. And one for the ball which is thrown upwards.
Let s1 represent the height of the dropped ball, above the ground
Let s2 represent the height of the thrown ball, above the ground.
Using the eqn of motion you have, S = -16t2 + v0t + s0, and substituting in appropriate values for s0 and v0, (upwards is taken as positive)
s1 = 60 - (1/2)gtē
s2 = 80t - (1/2)gtē
subtracting one height from t'other,
s1 - s2 = 60 - 80t
For the difference in height to be less than or equal to 10, we can write,
|s1 - s2| = |60 - 80t| <= 10
We therefore have two inequalities,
60 - 80t <= 10
and
80t - 60 <= 10
From the 1st inequality,
60 - 80t <= 10
60 - 10 <= 80t
80t >= 50
t >= 5/8
========
From the 2nd inequality,
80t - 60 <= 10
80t <= 70
t <= 7/8
========
The balls will be less than or equal to 10 feet apart in the time interval denoted by.
5/8 <= t <= 7/8
===============
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! 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.
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