GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation h(t) = –16t2 + 60t. At what times will the object be at a height of 56 feet?
A. 14 s, 6 s
B. 1.75 s, 2 s
C. 8 s, 1 s
D. −1.75 s, −2 s

The times that the object will be at a height of 56 feet is t = 1.75s and 2s

Given the formula that expresses the height as shown:

h(t) = -16t^2 + 60t

In order to geet the time, it will take the object be at a height of 56 feet, we will substitute h(t) = 56 into the expression to have:

56 =  -16t^2 + 60t

-16t^2 + 60t - 56 = 0

16t^2 - 60t + 56 = 0

Divide through by 4

4t^2 - 15t + 14 = 0

Factorize the equation

4t^2 - 15t + 14 = 0

t = 1.75s and 2s

Hence the times that the object will be at a height of 56 feet is t = 1.75s and 2s

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Solution:

Given that,

The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:

At what times will the object be at a height of 56 feet

Substitute h = 56

Solve the above equation by quadratic formula

Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

f-1(x)= cube root of 19x squared/x

step-by-step explanation:

b

2

−

b

b

3

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7

b

2

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14

b

+

6

%

−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation: