Question

The height in feet of an object propelled from a hill in t seconds is given
by h(t)= -16t² + 45t +400. Calculate the following. Round answers to the hundredths.
a) At what times is the object 200 ft high?
b) What height is the object 2 seconds after it propelled?
c) At what time does it reach the highest point?
d) What is the highest point the object reaches

Answer 1

Given :-

• Height in feet after t seconds h(t) = -16t² + 45t +400.

Question ❶ :-

• At what times is the object 200 ft high ?

Solution :-

To find the time when the altitude is 200ft Hight, Put h(t) = 200 .

→ h(t) = -16t² + 45t +400.

→ 200 = -16t² + 45t +400.

→ 16t² - 45t - 200 = 0

Now,

→ Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,

→ x = [ -b ± √(b²-4ac) / 2a ]

or,

→ x = [ - b ± √D /2a ] where D(Discriminant)= b²-4ac.

we Have :-

→ a = 16

→ b = (-45)

→ c = (-200)

Putting All values we get :-

→ t = [ 45 ± √{(-45)² - 4*16*(-200)} ] / 2*16

→ t = [ 45 ± √{ 2025 + 12800} ] / 32

→ t = [ 45 ± 121.75 ] / 32

→ t = (166.75)/32 = 5.2 seconds. { Negative value of Time Not Possible .}

Hence, After 5.2 Seconds the object is 200ft High.

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Question ❷ :-

• What height is the object 2 seconds after it propelled ?

Solution :-

Putting t = 2 , we get :-

→ h(t) = -16t² + 45t +400.

→ h(2) = -16(2)² + 45*2 + 400

→ h(2) = -64 + 90 + 400

→ h(2) = 490 - 64

→ h(2) = 426 ft. (Ans).

__________________________

Question ❸ :-

• At what time does it reach the highest point ?

Solution :-

→ h(t) = -16t² + 45t +400.

For hightest Point we have to make the Equation a perfect Square by using Completing The Square Method .

Taking (-16) common , or we can say that, dividing the Whole Equation by coefficient of x², we get,

→ h(t) = (-16)[ t² - (45/16)t - (400/16) ]

→ h(t) = (-16) [ t² - (45/16)t = (400/16) ]

Now, Adding And Subtracting (b/2a)² in LHS , we get,

→ h(t) = (-16) [ t² - (45/16)t - (2025/1024) + (2025/1024) - (400/16) ]

now,

→ 2ab = (45/16)t

→ 2tb = (45/16)t

→ b = (45/32)

So,

→ h(t) = (-16) [ t - (45/16) ]² + (2025/1024) - (400/16)

→ h(t) = (-16) [ t - (45/16) ]² + 431.64

So,

→ (t - (45/16)) = 0

→ t = (45/16)

→ t = 1.4 seconds. (Ans).

Hence, The ball reaches the highest point after 1.4 seconds.

Also,

Answer ❹ :-

The highest point the object reaches is 431.64ft.

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Answer 2

Answer:

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