Question 66214: I NEED HELP PLEASE!
Suppose a baseball is shot up from the ground straight up with an initial 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.
What is the function that describes this problem?
The ball will be how high above the ground after 1 second?
How long will it take to hit the ground?
What is the maximum height of the ball?
Found 2 solutions by venugopalramana, Earlsdon:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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 �g, 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).
---------
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.
Answer by Earlsdon(6294)  (Show Source):
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