SOLUTION: 18. 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

Question 49300: 18. 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 + vt + s0
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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
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Answer 17287 by venugopalramana(1619) 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)
ATTACHMENT:-
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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