Question 35905: Please help.
3) Suppose you throw a baseball straight up at a velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
• 16 represents ½g, the gravitational pull due to gravity (measured in feet per second 2).
• v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
• s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
a) What is the function that describes this problem?
Answer:
b) The ball will be how high above the ground after 1 second?
Answer:
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c) How long will it take to hit the ground?
Answer:
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d) What is the maximum height of the ball?
Answer:
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Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! OK .DONT WORRY LET ME SHOW YOU THE WORKING OF YOUR SPECIFIC PROBLEM
3) Suppose you throw a baseball straight up at a velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
• 16 represents ½g, the gravitational pull due to gravity (measured in feet per second 2).
• v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
• s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
a) What is the function that describes this problem?
Answer:
THIS IS WHAT IS GIVEN ABOVE
S=-16*T^2+V0*T+S0 .....................................I
WITH THE ELABORATIONS GIVEN ABOVE.
b) The ball will be how high above the ground after 1 second?
Answer:
Show work in this space.
WE ARE GIVEN
V0=32 FPS
S0=0...BALL THROWN FROM GROUND AND DISTANCE MEASURED FROM GROUND
T=1 SEC...
S=?
HENCE SUBSTITUTING IN EQN.I,WE GET
S=-16*1^2+32*1+0=-16+32=16 FT.
c) How long will it take to hit the ground?
Answer:
Show work in this space.
WE ARE GIVEN HERE
V0=32
S0=0
S=0
T=?
HENCE SUBSTITUTING IN EQN.I,WE GET
0=-16T^2+32T+0
16T(-T+2)=0
T=0...OR....-T+2=0....OR....T=2 SEC.
SINCE T=0 REPRESENTS THE INTIAL POSITION
T=2 SEC.IS THE ANSWER WHEN IT HITS THE GROUND AGAIN AFTER GOING UP AND FALLING DOWN.
d) What is the maximum height of the ball?
Answer:
Show work in this space.
HERE WE ARE GIVEN
V0=32
S0=0....AND THERE ARE 2 ASPECTS TO TAKE NOTE OF.
1.THE BALL GOES UP FIRST SLOWING DOWN AS IT GOES UP DUE TO EARTHS GRAVITATION...NOTE -16 GRAVITATIONAL ACCELERATIONED MENTIONED IN THE PROBLEM. IT GOES UP TILL ITS VELOCITY BECOMES ZERO FROM THE INITIAL VELOCITY OF THROW OF 32 FPS.AND THEN FALLS DOWN REGAINING THE SAME VELOCITY WHEN IT HITS THE GROUND AS PER PHYSICS LAWS,NEGLECTING AIR DRAG.
2.IT IS ALSO PROVED IN PHYSICS IN SUCH CASES,THAT
i)THE TIME OF ASCENT = THE TIME OF DESCENT
ii)THE DISTANCE TRAVELLED UPWARD=THE DISTANCE TRVELLED DOWN WARD
iii)THE VELOCITY WITH WHICH IT IS THROWN UP= THE VELOCITY WITH WHICH IT HITS THE GROUND.
HENCE USING i)PRINCIPLE ,SINCE TOTAL TIME OF TRAVEL AS PER C) AS WE ALCULATED IS 2 SEC.,TIME OF ASCENT =1 SEC.HENCE MAXIMUM HEIGHT REACHED IS DISTANCE TRAVELLED IN 1 SEC=16 FT.AS SHOWN IN B).
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ANOTHER METHOD IS TO FIND WHEN THE VELOCITY WILL BECOME ZERO AS IT GOES UP AND FIND THE DISTANCE TRAVELLED IN THAT TIME.
THE FORMULA FOR THAT IS OBTAINED BY DIFFERENTIATING THE GIVEN EQN.II
AS FOLLOWS
DS/DT=VELOCITY=V=-32*T+V0...WHERE V IS THE FINAL VELOCITY AFTER T SECS.
SINCE AT MAXMUM HEIGHT FINAL VELOCITY =0 (AS THEN ONLY IT FALLS BACK TOWARDS GROUND.)
HENCE 0=-32T+32
32T=32
T=1...AS WE USED EARLIER NOW WE FIND S=16 FT.AS BEFORE.
HOPE YOU UNDERSTOOD.
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