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Sagot :
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]
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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