Answer :
Certainly! Let's delve into each part of the question step-by-step.
a) From what height was the rocket launched?
To determine the height from which the rocket was launched, we need to evaluate the height function \( h(t) \) at \( t = 0 \). This gives us the y-intercept of the parabola.
[tex]\[ h(0) = -4.9(0)^2 + 1000(0) + 550 = 550 \][/tex]
So, the rocket was launched from a height of 550 meters above sea level.
b) What is the maximum height the rocket reaches?
To find the maximum height, we need to find the vertex of the parabola described by the function \( h(t) = -4.9 t^2 + 1000 t + 550 \). The vertex form for a quadratic function \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \).
Here, \( a = -4.9 \) and \( b = 1000 \), so:
[tex]\[ t = -\frac{1000}{2 \cdot (-4.9)} = \frac{1000}{9.8} = 102.04081632653061 \][/tex]
Now, substituting \( t = 102.04081632653061 \) back into the height equation \( h(t) \):
[tex]\[ h(102.04081632653061) = -4.9 (102.04081632653061)^2 + 1000 (102.04081632653061) + 550 = 51570.4081632653 \][/tex]
Therefore, the maximum height the rocket reaches is 51570.408 meters.
c) If the rocket will splash down in the ocean, when will it splash down?
The rocket will splash down in the ocean when its height \( h(t) \) is zero. To find this, we solve the quadratic equation \( -4.9 t^2 + 1000 t + 550 = 0 \).
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = -4.9 \), \( b = 1000 \), and \( c = 550 \).
The solutions are:
[tex]\[ t = \frac{-1000 \pm \sqrt{1000^2 - 4(-4.9)(550)}}{2(-4.9)} \][/tex]
Solving this will yield two solutions, one of which is negative and irrelevant in the context (we cannot have negative time). Hence, we consider the positive solution:
[tex]\[ t = 204.630158338964 \][/tex]
Therefore, the rocket will splash down in the ocean at approximately 204.63 seconds after launch.
a) From what height was the rocket launched?
To determine the height from which the rocket was launched, we need to evaluate the height function \( h(t) \) at \( t = 0 \). This gives us the y-intercept of the parabola.
[tex]\[ h(0) = -4.9(0)^2 + 1000(0) + 550 = 550 \][/tex]
So, the rocket was launched from a height of 550 meters above sea level.
b) What is the maximum height the rocket reaches?
To find the maximum height, we need to find the vertex of the parabola described by the function \( h(t) = -4.9 t^2 + 1000 t + 550 \). The vertex form for a quadratic function \( at^2 + bt + c \) occurs at \( t = -\frac{b}{2a} \).
Here, \( a = -4.9 \) and \( b = 1000 \), so:
[tex]\[ t = -\frac{1000}{2 \cdot (-4.9)} = \frac{1000}{9.8} = 102.04081632653061 \][/tex]
Now, substituting \( t = 102.04081632653061 \) back into the height equation \( h(t) \):
[tex]\[ h(102.04081632653061) = -4.9 (102.04081632653061)^2 + 1000 (102.04081632653061) + 550 = 51570.4081632653 \][/tex]
Therefore, the maximum height the rocket reaches is 51570.408 meters.
c) If the rocket will splash down in the ocean, when will it splash down?
The rocket will splash down in the ocean when its height \( h(t) \) is zero. To find this, we solve the quadratic equation \( -4.9 t^2 + 1000 t + 550 = 0 \).
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = -4.9 \), \( b = 1000 \), and \( c = 550 \).
The solutions are:
[tex]\[ t = \frac{-1000 \pm \sqrt{1000^2 - 4(-4.9)(550)}}{2(-4.9)} \][/tex]
Solving this will yield two solutions, one of which is negative and irrelevant in the context (we cannot have negative time). Hence, we consider the positive solution:
[tex]\[ t = 204.630158338964 \][/tex]
Therefore, the rocket will splash down in the ocean at approximately 204.63 seconds after launch.
