Answer :
To address the given quadratic function [tex]\( h = -4.9t^2 + 50t + 30 \)[/tex] which models the height [tex]\( h \)[/tex] of a rocket over time [tex]\( t \)[/tex], we'll go through each part of the question step-by-step.
### (a) Graph the height versus time for the rocket
To graph the height [tex]\( h \)[/tex] against time [tex]\( t \)[/tex], you can plot the equation [tex]\( h = -4.9t^2 + 50t + 30 \)[/tex]. This is a quadratic function that forms a parabola opening downwards because the coefficient of [tex]\( t^2 \)[/tex] is negative.
1. At [tex]\( t = 0 \)[/tex] (launch time), the height can be calculated as:
[tex]\[ h = -4.9(0)^2 + 50(0) + 30 = 30 \text{ meters} \][/tex]
2. Plot other points by picking values of [tex]\( t \)[/tex] within a reasonable range (e.g., 0 to 12 seconds). Use a graphing calculator or an online tool like Desmos to visualize it. Remember to mark the important points such as the vertex and when the rocket hits the ground.
### (b) Determine and label the height of the rocket at its vertex maximum
For a quadratic function in standard form [tex]\( h = at^2 + bt + c \)[/tex], the time at which the vertex occurs is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -4.9 \)[/tex] and [tex]\( b = 50 \)[/tex]:
[tex]\[ t_{\text{vertex}} = -\frac{50}{2 \times -4.9} = 5.102 \text{ seconds (approx)} \][/tex]
Next, calculate the height at this time by plugging [tex]\( t_{\text{vertex}} \)[/tex] back into the function:
[tex]\[ h_{\text{vertex}} = -4.9 \times (5.102)^2 + 50 \times 5.102 + 30 \approx 157.55 \text{ meters} \][/tex]
Label this point on the graph at approximately [tex]\( (5.102, 157.551) \)[/tex].
### (c) Determine and label the coordinate that shows how many seconds it will take before the rocket hits the ground
To find when the rocket hits the ground, solve for [tex]\( t \)[/tex] when [tex]\( h = 0 \)[/tex]:
[tex]\[ -4.9t^2 + 50t + 30 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], which can be solved using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Calculate the discriminant first:
[tex]\[ \Delta = b^2 - 4ac = 50^2 - 4(-4.9)(30) = 2500 + 588 = 3088 \][/tex]
Then solve for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{ -50 + \sqrt{3088} }{2(-4.9)} = -0.568 \text{ seconds (approx, non-physical as time cannot be negative)} \][/tex]
[tex]\[ t_2 = \frac{ -50 - \sqrt{3088} }{2(-4.9)} = 10.772 \text{ seconds (approx)} \][/tex]
Since negative time doesn't make physical sense here, the time at which the rocket hits the ground is approximately [tex]\( 10.772 \)[/tex] seconds.
Label this point on the graph at approximately [tex]\( (10.772, 0) \)[/tex].
By carefully following these steps, you can effectively interpret and label the important points on the graph of the rocket's height over time. The vertex maximum occurs at [tex]\( t \approx 5.102 \)[/tex] seconds with a height of approximately [tex]\( 157.551 \)[/tex] meters, and the rocket hits the ground at approximately [tex]\( t \approx 10.772 \)[/tex] seconds.
### (a) Graph the height versus time for the rocket
To graph the height [tex]\( h \)[/tex] against time [tex]\( t \)[/tex], you can plot the equation [tex]\( h = -4.9t^2 + 50t + 30 \)[/tex]. This is a quadratic function that forms a parabola opening downwards because the coefficient of [tex]\( t^2 \)[/tex] is negative.
1. At [tex]\( t = 0 \)[/tex] (launch time), the height can be calculated as:
[tex]\[ h = -4.9(0)^2 + 50(0) + 30 = 30 \text{ meters} \][/tex]
2. Plot other points by picking values of [tex]\( t \)[/tex] within a reasonable range (e.g., 0 to 12 seconds). Use a graphing calculator or an online tool like Desmos to visualize it. Remember to mark the important points such as the vertex and when the rocket hits the ground.
### (b) Determine and label the height of the rocket at its vertex maximum
For a quadratic function in standard form [tex]\( h = at^2 + bt + c \)[/tex], the time at which the vertex occurs is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -4.9 \)[/tex] and [tex]\( b = 50 \)[/tex]:
[tex]\[ t_{\text{vertex}} = -\frac{50}{2 \times -4.9} = 5.102 \text{ seconds (approx)} \][/tex]
Next, calculate the height at this time by plugging [tex]\( t_{\text{vertex}} \)[/tex] back into the function:
[tex]\[ h_{\text{vertex}} = -4.9 \times (5.102)^2 + 50 \times 5.102 + 30 \approx 157.55 \text{ meters} \][/tex]
Label this point on the graph at approximately [tex]\( (5.102, 157.551) \)[/tex].
### (c) Determine and label the coordinate that shows how many seconds it will take before the rocket hits the ground
To find when the rocket hits the ground, solve for [tex]\( t \)[/tex] when [tex]\( h = 0 \)[/tex]:
[tex]\[ -4.9t^2 + 50t + 30 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], which can be solved using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Calculate the discriminant first:
[tex]\[ \Delta = b^2 - 4ac = 50^2 - 4(-4.9)(30) = 2500 + 588 = 3088 \][/tex]
Then solve for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{ -50 + \sqrt{3088} }{2(-4.9)} = -0.568 \text{ seconds (approx, non-physical as time cannot be negative)} \][/tex]
[tex]\[ t_2 = \frac{ -50 - \sqrt{3088} }{2(-4.9)} = 10.772 \text{ seconds (approx)} \][/tex]
Since negative time doesn't make physical sense here, the time at which the rocket hits the ground is approximately [tex]\( 10.772 \)[/tex] seconds.
Label this point on the graph at approximately [tex]\( (10.772, 0) \)[/tex].
By carefully following these steps, you can effectively interpret and label the important points on the graph of the rocket's height over time. The vertex maximum occurs at [tex]\( t \approx 5.102 \)[/tex] seconds with a height of approximately [tex]\( 157.551 \)[/tex] meters, and the rocket hits the ground at approximately [tex]\( t \approx 10.772 \)[/tex] seconds.
