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John hits the ball at 3 ft above the ground with an initial velocity of 32 ft/sec.The path of the ball is given by the function S (t) = 16t² +32 +3 .where t is the time in seconds and S is the height.what is the maximum height reached the ball ?
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step by step solution ..​

Sagot :

Hurgfsidn

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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