If an object is dropped from a height of 55 feet, the function d = - 16^2 + 55 gives the height of the object after t seconds. Graph this function. Approximately how long does it take the object to reach the ground (d=0)

t is approximately 1.854049622 seconds

Step-by-step explanation:

d = - 16 t^2 + 55

Let d = 0

0 = - 16 t^2 + 55

Subtract 55 from each side

-55 = - 16 t^2

Divide by - 16 on each side

-55/-16 = - 16 / -16t^2

55/16 = t^2

Take the square root of each side

sqrt (55/16) = sqrt (t^2)

We only take the positive square root because time must be positive

sqrt (55/16) = t

t is approximately 1.854049622 seconds

Approximately 1.9 seconds (correct to nearest tenth)

Step-by-step explanation:

Looks like the function is d = - 16t^2 + 55 (you left out the t)

The answer is the value of t when d = 0 so we have the equation:-

0 = - 16t^2 + 55

16t^2 = 55

t^2 = 55/16

t = sqrt (55/16)

= 1.85 seconds