Answer :
To solve for the time at which the object will reach the ground, \(t\), we need to find the value of \(t\) when the height \(h(t)\) is zero. This scenario describes when the object hits the ground.
Given the height equation:
[tex]\[h(t) = -16t^2 + 45t + 75\][/tex]
We need to find \(t\) when \(h(t) = 0\):
[tex]\[0 = -16t^2 + 45t + 75\][/tex]
This is a quadratic equation in the standard form [tex]\[ax^2 + bx + c = 0\][/tex], where \(a = -16\), \(b = 45\), and \(c = 75\).
The quadratic formula to solve for \(t\) is given by:
[tex]\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Let's start by calculating the discriminant, which is the part under the square root:
[tex]\[ \text{discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{discriminant} = 45^2 - 4(-16)(75) \][/tex]
[tex]\[ \text{discriminant} = 2025 + 4800 \][/tex]
[tex]\[ \text{discriminant} = 6825 \][/tex]
Now we take the square root of the discriminant:
[tex]\[ \sqrt{6825} \approx 82.6221 \][/tex]
Next, we can find the two potential solutions for \(t\):
[tex]\[ t_1 = \frac{-b + \sqrt{\text{discriminant}}}{2a} \][/tex]
[tex]\[ t_1 = \frac{-45 + 82.6221}{-32} \][/tex]
[tex]\[ t_1 = \frac{37.6221}{-32} \][/tex]
[tex]\[ t_1 \approx -1.176 \][/tex]
[tex]\[ t_2 = \frac{-b - \sqrt{\text{discriminant}}}{2a} \][/tex]
[tex]\[ t_2 = \frac{-45 - 82.6221}{-32} \][/tex]
[tex]\[ t_2 = \frac{-127.6221}{-32} \][/tex]
[tex]\[ t_2 \approx 3.988 \][/tex]
Therefore, the solutions to the equation \( -16t^2 + 45t + 75 = 0 \) are approximately \( t_1 \approx -1.176 \, \text{seconds} \) and \( t_2 \approx 3.988 \, \text{seconds} \).
Since time cannot be negative, we discard \( t_1 \approx -1.176 \).
Thus, the time at which the object will hit the ground is approximately [tex]\( t \approx 3.988 \, \text{seconds} \)[/tex].
Given the height equation:
[tex]\[h(t) = -16t^2 + 45t + 75\][/tex]
We need to find \(t\) when \(h(t) = 0\):
[tex]\[0 = -16t^2 + 45t + 75\][/tex]
This is a quadratic equation in the standard form [tex]\[ax^2 + bx + c = 0\][/tex], where \(a = -16\), \(b = 45\), and \(c = 75\).
The quadratic formula to solve for \(t\) is given by:
[tex]\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Let's start by calculating the discriminant, which is the part under the square root:
[tex]\[ \text{discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{discriminant} = 45^2 - 4(-16)(75) \][/tex]
[tex]\[ \text{discriminant} = 2025 + 4800 \][/tex]
[tex]\[ \text{discriminant} = 6825 \][/tex]
Now we take the square root of the discriminant:
[tex]\[ \sqrt{6825} \approx 82.6221 \][/tex]
Next, we can find the two potential solutions for \(t\):
[tex]\[ t_1 = \frac{-b + \sqrt{\text{discriminant}}}{2a} \][/tex]
[tex]\[ t_1 = \frac{-45 + 82.6221}{-32} \][/tex]
[tex]\[ t_1 = \frac{37.6221}{-32} \][/tex]
[tex]\[ t_1 \approx -1.176 \][/tex]
[tex]\[ t_2 = \frac{-b - \sqrt{\text{discriminant}}}{2a} \][/tex]
[tex]\[ t_2 = \frac{-45 - 82.6221}{-32} \][/tex]
[tex]\[ t_2 = \frac{-127.6221}{-32} \][/tex]
[tex]\[ t_2 \approx 3.988 \][/tex]
Therefore, the solutions to the equation \( -16t^2 + 45t + 75 = 0 \) are approximately \( t_1 \approx -1.176 \, \text{seconds} \) and \( t_2 \approx 3.988 \, \text{seconds} \).
Since time cannot be negative, we discard \( t_1 \approx -1.176 \).
Thus, the time at which the object will hit the ground is approximately [tex]\( t \approx 3.988 \, \text{seconds} \)[/tex].
