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A ball is shot from a cannon into the air with an upward velocity of [tex][tex]$40 \, \text{ft/sec}$[/tex][/tex]. The function that gives the height [tex](h)[/tex] of the ball at any time [tex](t)[/tex] is [tex]h(t) = -16t^2 + 40t + 1.5[/tex].

1. Find the maximum height attained by the ball.
2. Determine how much time has passed when the cannonball hits the ground.

Round both answers to the nearest tenth.

The maximum height of the cannonball is [tex]\square[/tex] feet.

The cannonball is in the air for [tex]\square[/tex] seconds.



Answer :

To solve the problem, we need to determine two key pieces of information from the function [tex]\( h(t) = -16t^2 + 40t + 1.5 \)[/tex]:

1. The maximum height the ball reaches.
2. The time it takes for the ball to hit the ground.

### Step 1: Finding the Maximum Height
The height function [tex]\( h(t) = -16t^2 + 40t + 1.5 \)[/tex] is a quadratic equation in standard form [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 40 \)[/tex], and [tex]\( c = 1.5 \)[/tex].

The maximum height of the ball, given by the parabola opening downwards (since [tex]\( a < 0 \)[/tex]), occurs at the vertex. The [tex]\( t \)[/tex]-coordinate of the vertex for a quadratic function [tex]\( at^2 + bt + c \)[/tex] is given by:
[tex]\[ t_{\text{max height}} = -\frac{b}{2a} \][/tex]

Plugging in the values:
[tex]\[ t_{\text{max height}} = -\frac{40}{2(-16)} = \frac{40}{32} = 1.25 \text{ seconds} \][/tex]

Next, we determine the height at this time by substituting [tex]\( t = 1.25 \)[/tex] back into the height function:
[tex]\[ h(1.25) = -16(1.25)^2 + 40(1.25) + 1.5 \][/tex]

Calculating this,
[tex]\[ h(1.25) = -16 \times 1.5625 + 40 \times 1.25 + 1.5 \][/tex]
[tex]\[ h(1.25) = -25 + 50 + 1.5 \][/tex]
[tex]\[ h(1.25) = 26.5 \text{ feet} \][/tex]

So, the maximum height attained by the ball is:
[tex]\[ 26.5 \text{ feet} \][/tex]

### Step 2: Finding the Time When the Ball Hits the Ground
The ball hits the ground when the height [tex]\( h(t) \)[/tex] is zero. Therefore, we set the height function equal to zero and solve for [tex]\( t \)[/tex].
[tex]\[ -16t^2 + 40t + 1.5 = 0 \][/tex]

Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{-40 \pm \sqrt{40^2 - 4(-16)(1.5)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-40 \pm \sqrt{1600 + 96}}{-32} \][/tex]
[tex]\[ t = \frac{-40 \pm \sqrt{1696}}{-32} \][/tex]
[tex]\[ t = \frac{-40 \pm 41.2}{-32} \][/tex]

We get two solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-40 + 41.2}{-32} = \frac{1.2}{-32} = -0.0375 \text{ seconds (not physically meaningful)} \][/tex]
[tex]\[ t_2 = \frac{-40 - 41.2}{-32} = \frac{-81.2}{-32} = 2.54 \text{ seconds} \][/tex]

Rounding to the nearest tenth:
[tex]\[ t = 2.5 \text{ seconds} \][/tex]

Therefore, the cannonball is in the air for:
[tex]\[ 2.5 \text{ seconds} \][/tex]

### Summary
1. The maximum height of the cannonball is:
[tex]\[ 26.5 \text{ feet} \][/tex]
2. The cannonball is in the air for:
[tex]\[ 2.5 \text{ seconds} \][/tex]

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