Question

solve using square roots.
an object is dropped from a height of 64 feet off the ground. the height $h$ of the object after $t$ seconds can be found using the equation $h = 64 - 16t^2$
when will the height be 48 feet?
$square$ second(s)
when will the object reach the ground?
$square$ second(s)

Explanation:

Step1: Set h=48, solve for t

Substitute $h=48$ into $h=64-16t^2$:
$$48=64-16t^2$$

Step2: Isolate the $t^2$ term

Rearrange to solve for $16t^2$:
$$16t^2=64-48$$
$$16t^2=16$$

Step3: Solve for $t^2$

Divide both sides by 16:
$$t^2=1$$

Step4: Take square root

$$t=\sqrt{1}=1$$
(We discard the negative root as time cannot be negative)

Step5: Set h=0, solve for t

Substitute $h=0$ into $h=64-16t^2$:
$$0=64-16t^2$$

Step6: Isolate the $t^2$ term

Rearrange to solve for $16t^2$:
$$16t^2=64$$

Step7: Solve for $t^2$

Divide both sides by 16:
$$t^2=4$$

Step8: Take square root

$$t=\sqrt{4}=2$$
(We discard the negative root as time cannot be negative)

Answer:

When will the height be 48 feet? 1 second(s)
When will the object reach the ground? 2 second(s)