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The height of an object dropped from the top of a 144-foot building is given by [tex]$h(t) = -16t^2 + 144$[/tex]. How long will it take the object to hit the ground?

Sagot :

To determine how long it will take for the object to hit the ground, we'll use the given height function:

[tex]\[ h(t) = -16t^2 + 144 \][/tex]

Here, [tex]\( h(t) \)[/tex] represents the height of the object at time [tex]\( t \)[/tex]. We need to find the time [tex]\( t \)[/tex] when the object hits the ground, which means [tex]\( h(t) = 0 \)[/tex].

Let's set the equation to zero and solve for [tex]\( t \)[/tex]:

[tex]\[ -16t^2 + 144 = 0 \][/tex]

First, move 144 to the other side of the equation:

[tex]\[ -16t^2 = -144 \][/tex]

Next, divide both sides of the equation by -16 to isolate [tex]\( t^2 \)[/tex]:

[tex]\[ t^2 = \frac{144}{16} \][/tex]

Simplify the fraction:

[tex]\[ t^2 = 9 \][/tex]

To find [tex]\( t \)[/tex], take the square root of both sides:

[tex]\[ t = \sqrt{9} \][/tex]

Thus, we have two potential values for [tex]\( t \)[/tex]:

[tex]\[ t = 3 \quad \text{or} \quad t = -3 \][/tex]

Since time cannot be negative in this context, we discard [tex]\( t = -3 \)[/tex]. Thus, the time [tex]\( t \)[/tex] when the object hits the ground is:

[tex]\[ t = 3 \text{ seconds} \][/tex]

So, it will take 3 seconds for the object to hit the ground.