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A ball is launched out of a cannon that is 24 feet off the ground. The height of the ball after "t" seconds is represented by
h(t) = -16t² + 20x + 24
How many seconds will it take for the to reach the ground?

Sagot :

Given:

The height of ball is represented by the below function:

[tex]h(t)=-16t^2+20t+24[/tex]

To find:

The number of seconds it will take to reach the ground.

Solution:

We have,

[tex]h(t)=-16t^2+20t+24[/tex]

At ground level, the height of ball is 0, i.e., [tex]h(t)=0[/tex].

[tex]-16t^2+20t+24=0[/tex]

Taking out greatest common factor.

[tex]-4(4t^2-5t-6)=0[/tex]

[tex]4t^2-5t-6=0[/tex]

Splitting the middle term, we get

[tex]4t^2-8t+3t-6=0[/tex]

[tex]4t(t-2)+3(t-2)=0[/tex]

[tex](t-2)(4t+3)=0[/tex]

Using zero product property, we get

[tex](t-2)=0[/tex] and [tex](4t+3)=0[/tex]

[tex]t=2[/tex] and [tex]t=-\dfrac{3}{4}[/tex]

Time cannot be negative, so [tex]t=2[/tex].

Therefore, the ball will reach the ground after 2 seconds.

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