Answer:
Maximum height of the ball will reached to a height of 19 units at t=1.
Step-by-step explanation:
A question where a typo error at S (t) = 16t² +32 +3,
Correcting the typo error
S (t) = 16t² +32t +3,
The maximum height reached the ball will occur at
t = -b/(2a)
Where
S(t) = at^2 + bt + c.
Thefore
a = 16
b = 32
Solving
t = -b/(2a)
t = -32/2(16)
t = -1
There is no negative time therefore the question has a typo error again
Solving further by substituting t = -1 to the function S(t)
S (t) = 16t² +32t +3,
S (-1) = 16(-1)² +32(-1) +3,
S (-1) = 16-32+3
S (-1) = -13
This is a further indication that the question has a typo error since there is no negative height
Correcting the typo error S (t) = 16t² +32t +3
S (t) = -16t² +32t +3
a = -16
b = 32
Substituting the values of a and b to t = -b/(2a)
t = -b/(2a)
t = -32/(2)(-16)
t = 1
Substitute the value of t to the function S(t)
S (t) = -16t² +32t +3
S(1) = -16(1)² + 32(1) + 3
S(1) = -16+32+3
S(1) = 19
Therefore the maximum height of the ball will reached to a height of 19 units at t=1.
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