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An object is launched straight into the air. The projectile motion of the object can be modeled using [tex]h(t)=96t-16t^2[/tex], where [tex]t[/tex] is the time since launch and [tex]h(t)[/tex] is the height in feet of the projectile after time [tex]t[/tex] in seconds.

a) When will the object be 144 feet in the air? [tex]\square[/tex]

b) When will the object hit the ground? [tex]\square[/tex]

Sagot :

To solve these problems, we'll start with the height function given by [tex]\( h(t) = 96t - 16t^2 \)[/tex], where [tex]\( h(t) \)[/tex] represents the height in feet and [tex]\( t \)[/tex] represents the time in seconds.

### Finding When the Object is 144 Feet in the Air

We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is 144 feet. Setting [tex]\( h(t) = 144 \)[/tex], we get the equation:
[tex]\[ 96t - 16t^2 = 144 \][/tex]

Rearranging this equation, we bring all terms to one side:
[tex]\[ 16t^2 - 96t + 144 = 0 \][/tex]

Next, we divide the entire equation by 16 to simplify:
[tex]\[ t^2 - 6t + 9 = 0 \][/tex]

This simplifies to a quadratic equation that can be solved by factoring:
[tex]\[ (t - 3)^2 = 0 \][/tex]

Solving for [tex]\( t \)[/tex], we get:
[tex]\[ t = 3 \][/tex]

Therefore, the object will be 144 feet in the air at [tex]\( t = 3 \)[/tex] seconds.

### Finding When the Object Hits the Ground

To determine when the object hits the ground, we set [tex]\( h(t) = 0 \)[/tex]:
[tex]\[ 96t - 16t^2 = 0 \][/tex]

We can factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t(96 - 16t) = 0 \][/tex]

Setting each factor to zero gives us:
[tex]\[ t = 0 \quad \text{or} \quad 96 - 16t = 0 \][/tex]

Solving the second equation for [tex]\( t \)[/tex]:
[tex]\[ 96 = 16t \\ t = \frac{96}{16} \\ t = 6 \][/tex]

Therefore, the object hits the ground at [tex]\( t = 6 \)[/tex] seconds.

### Summary

- The object will be 144 feet in the air at [tex]\( t = 3 \)[/tex] seconds.
- The object will hit the ground at [tex]\( t = 0 \)[/tex] seconds (launch time) and [tex]\( t = 6 \)[/tex] seconds.