154k views
0 votes
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)

2 Answers

4 votes

Answer:

Approximately 1.9 seconds (correct to nearest tenth)

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

User SiddharthaRT
by
8.3k points
5 votes

Answer:

t is approximately 1.854049622 seconds

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

If an object is dropped from a height of 55 feet, the function d = -16^2 + 55 gives-example-1
User Texasflood
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories