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Xelga [282]
3 years ago
13

Suppose you hit a fly ball with an initial upward velocity of 20 feet per second. Which of the following equations would be a re

alistic model for the height of the ball after t seconds?
Mathematics
2 answers:
ozzi3 years ago
8 0
\bf \qquad \textit{initial velocity}\\\\\begin{array}{llll}\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}
\end{cases}\\\\
-------------------------------\\\\
\textit{"initial upward velocity of 20 feet per second" simply means } v_o=20
baherus [9]3 years ago
3 0
A=g≈-32

v=⌠a dt  since a≈-32

v=-32t+C, where C is the initial velocity, which we are told is 20 ft/s

v=-32t+20

h=⌠v dt

h=-32t^2/2+20t+C, where C is the initial height, so what is a reasonable initial height?  How about 3 ft since we are swinging a bat around maybe our waist level... maybe :P

h=-16t^2+20t+3

Your choices may have made a different assumption about the initial height of course but you did not show your choices, but certainly it will be:

h(t)=-16t^2+20t+hi, where hi is the initial height in feet.

This of course ignores air resistance and flight dynamics of a spinning ball with seams, which will make a significant difference in real life :)
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Consider the following game, played with three standard six-sided dice. If the player ends with all three dice showing the same
ddd [48]

Answer:

a

 P(A) =  \frac{1}{36}

b

P(B) = \frac{15}{36}

c

P(U) = \frac{11}{36}

Step-by-step explanation:

From the question we are told that

   The  number of dice is  n  =  3

Generally  for all three dice to show the same number the second and the third dice must have the same outcome  as the outcome of the first dice.

This means that the number of outcome the first die can  have is  6  (6 sides), the number of outcome which the second and the third dice can have is  1 (they must match the first )

So the probability that all three dice show the same number on the first roll is mathematically represented as

      P(A) =  \frac{6}{6} * \frac{1}{6} *  \frac{1}{6}

=>   P(A) =  \frac{1}{36}

Generally for two of the three dice show the same number after the first roll then the number of outcome for  one of the  dice would be is  6 , the number of outcome for another one must be   1(i.e it must match the first ) , the number of outcome for the remaining one must be   5 (i.e it can show any of the remaining  5  sides which the first and second dice are not showing )

Now the number of ways of selecting this 2 dice that show the same number from the 3 dice is mathematically represented as

     N  =  ^3C_2

Here C stands for combination

So

      N  =  ^3C_2  = 6

So the probability that exactly two of the three dice show the same number after the first roll is mathematically represented as  

      P(B) = N   \frac{1}{6} *  \frac{5}{6}

=>  P(B) = \frac{15}{36}

Generally from the question we are told that if two dice match, the player re-rolls the die that does not match.

Now the probability that  the die that did not match the first  time will match the second time is

     P(E ) =  \frac{1}{6}

Generally if that one die does not show the same number in the second round , the probability that it will match  in the third round is

    P(R) =  \frac{5}{6} *  \frac{1}{6}

Generally the probability that he wins (i.e when all three are showing the same number ) and  exactly two is showing the same number is mathematically represented as

       P(K) =  P(E)* P(B) + P(R)* P(B)

=>    P(K) =  \frac{1}{6} * \frac{15}{36}  +  \frac{5}{6} *  \frac{1}{6} * \frac{15}{36}

=>  P(K)= \frac{165}{1296}

So the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll is mathematically represented as  

    P(U) =  \frac{\frac{165}{1296}}{\frac{15}{36} }

=>  P(U) = \frac{11}{36}

     

 

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MakcuM [25]

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3.6/ x for x =2

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