A model rocket is launched from a roof of a building. Its path is described by the relationship h=-5t*2+35t+10, where h is the height in metres and t is the time in seconds. How tall is the building? What is the maximum height that the rocket will reach? When will the rocket be 40 metres above the ground? Explain your answer.
The building is 10 metres tall.
h=-5t^2+35t+10
=-5(t^2-7t-2)
=-5(t^2-7t+12.25-12.25)+10
=-5(t^2-7t+12.25)+61.25+10
= -5(t^2-7t+12.25)+71.25
=-5(t-3.5)^2 + 71.25
So at 3.5 seconds the maximum height is 71.25 metres high.
Is this correct?
Are the first 2 parts correct?
Hint please on the third part? I think I need the velocity? How to calculate?
To determine the height of the building, we need to find the value of "h" when "t" is zero. Substituting t=0 into the equation, we get:
h = -5(0)^2 + 35(0) + 10
h = 0 + 0 + 10
h = 10
Therefore, the height of the building is 10 meters.
To find the maximum height that the rocket will reach, we need to find the vertex of the quadratic equation. The vertex can be determined using the formula -b/2a, where a=-5 and b=35. Substituting these values, we get:
t = -35 / (2 * -5)
t = -35 / -10
t = 3.5
Substituting t=3.5 back into the equation, we can find the maximum height:
h = -5(3.5)^2 + 35(3.5) + 10
h = -5(12.25) + 122.5 + 10
h = -61.25 + 122.5 + 10
h = 71.25
Therefore, the maximum height that the rocket will reach is 71.25 meters.
To find when the rocket will be 40 meters above the ground, we need to set h=40 and solve for t. Rearranging the equation, we have:
-5t^2 + 35t + 10 = 40
-5t^2 + 35t - 30 = 0
Factoring the quadratic equation, we get:
(-5t - 2)(t - 15) = 0
Setting each factor equal to zero and solving for t, we get:
-5t - 2 = 0, t = -2/5
t - 15 = 0, t = 15
Since time cannot be negative in this context, we ignore t = -2/5. Therefore, the rocket will be 40 meters above the ground when t = 15 seconds.
The rocket will be 40 meters above the ground at 15 seconds because at that point, the height of the rocket, h, will be equal to 40.
To find the height of the building, we need to find the value of h when t is equal to zero.
Given the equation h = -5t^2 + 35t + 10, we substitute t = 0:
h = -5(0)^2 + 35(0) + 10
h = 0 + 0 + 10
h = 10
Therefore, the height of the building is 10 meters.
To find the maximum height that the rocket will reach, we need to find the vertex or the highest point on the graph of the equation.
The equation h = -5t^2 + 35t + 10 is in the form of a quadratic equation h = at^2 + bt + c, where a = -5, b = 35, and c = 10.
The equation for the vertex (max height) is t = -b/2a.
Substituting the values of a and b:
t = -35 / (2 * -5)
t = -35 / -10
t = 3.5
To find the height at t = 3.5, we substitute it into the equation:
h = -5(3.5)^2 + 35(3.5) + 10
h = -5(12.25) + 122.5 + 10
h = -61.25 + 122.5 + 10
h = 71.25
Therefore, the maximum height the rocket will reach is 71.25 meters.
To find when the rocket will be 40 meters above the ground, we set h equal to 40 and solve for t.
40 = -5t^2 + 35t + 10
Rearranging the equation:
5t^2 - 35t + 10 - 40 = 0
5t^2 - 35t - 30 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values a = 5, b = -35, and c = -30:
t = (-(-35) ± √((-35)^2 - 4(5)(-30))) / (2 * 5)
t = (35 ± √(1225 + 600)) / 10
t = (35 ± √(1825)) / 10
t = (35 ± √(5 * 365)) / 10
t = (35 ± 5√(73)) / 10
This gives us two possible solutions for t. However, we are only interested in positive values of t since time cannot be negative in this context. Therefore, we take the positive solution:
t = (35 + 5√(73)) / 10
This is the time at which the rocket will be 40 meters above the ground.
In conclusion, the height of the building is 10 meters, the maximum height that the rocket will reach is 71.25 meters, and the rocket will be 40 meters above the ground at the time (35 + 5√(73)) / 10 seconds.