A model rocket is launched from the roof of a building. It's flight path is modeled by -5t^2 + 30t + 10

Question

where h is the height of the rocket above the ground in meters and t is the time after the launch in seconds.

What is the rocket height after 2 seconds? 

At what time does the rocket reach its maximum height? How far is it above the ground at this time? 

 At what time does the rocket hit the ground?

Answer 1

Hi Valeria,

The equation they gave you is a quadratic equation. Before we solve the problem, it helps to understand the equation and graph of quadratic equations.

All quadratic equations have the form of:

y = ax² + bx + c

And all quadratic equations have the form of a parabola, which looks like a "U" when graphed.

In the standard quadratic equation above, the leading coefficient, a, tells us whether the parabola opens up (it looks like a "U") or opens down (it looks like an upside-down "U"). If a is positive (a > 0), then the parabola opens up; if a is negative (a < 0), then the parabola opens down.

Now let's look at the equation they gave us:

h = -5*t² + 30*t + 10

In this equation, h is the height of the rocket in meters at any given time, t, which is in seconds. All we did was replace x with t and y with h. If we correlate the coefficients of each term in our equation to the coefficients a, b and c from the standard equation, we see that a = -5, b = 30 and c = 10.

The leading coefficient, a, in our equation is negative, so our parabola opens down. This makes sense since a rocket that is launched into the air will first go up, but -- as gravity pulls it down to the ground -- then reach a maximum height before falling to the ground. If we were to draw the rocket's path, it would like a parabola that opens down:

maximum height

*

* *

* *

* *

roof --------- *

| *

| *

|------------------------------ ground

Let's answer the questions:

To find the rocket's height at t = 2 sec, we just have to substitute 2 into the equation for t and then solve for h:

h = -5*t² + 30*t + 10

h = -5(2)² + 30(2) + 10

h = -5(4) + 30(2) + 10

h = -20 + 60 +10

h = 50 m

The height of the rocket 2 seconds after launching it is 50 meters.

The maximum height of the rocket, h_max, and the time that the rocket reaches this maximum height, t_max, can be visualized from the graph as the vertex ( t_max , h_max ) of the parabola. In a parabola that opens up, the vertex is the minimum (or lowest) value of the function, but in a parabola that opens down (like ours) the vertex is the maximum (or greatest) value of the function. To find the x-coordinate of the vertex of a parabola, we use the formula:

x_vertex = -b / 2a

The x-coordinate in our equation is just t. So, to find the time that the rocket reaches its maximum height, we just solve for t_max:

t_max = -b / 2a

t_max = -(30) / 2(-5)

t_max = (-30) / (-10)

t_max = 3 sec

Our rocket reaches its maximum height at 3 seconds. To find the maximum height, h_max, we just substitute t_max = 3 into our quadratic equation and solve for h_max:

h_max = -5*t_max² + 30*t_max + 10

h_max = -5(3)² + 30(3) + 10

h_max = -5(9) + 30(3) + 10

h_max = -45 + 90 + 10

h_max = 55 m

Our rocket's maximum height is 55 meters.

Finally, we want to find the time at which the rocket hits the ground. Because the height of the rocket when it's on the ground is 0 meters, what we are actually looking for is the x-intercept, or the point at which the parabola crosses the x-axis. To find when the rockets hits the ground, we just substitute 0 into our equation for h and solve for t:

h = -5*t² + 30*t + 10

0 = -5(t² - 6*t - 2)

0 = t² - 6*t - 2

Because this quadratic polynomial cannot be factored, we have to use the quadratic formula or completing the square method to solve for t. I'm going to use completing the square:

t² - 6*t - 2 = 0

t² - 6*t = 2

t² - 6*t + 9 = 2 + 9

(t - 3)² = 11

t - 3 = ± √(11)

t = 3 ± √(11)

t = 3 ± 3.3

Solving for t, we get two solutions:

t = 3 - 3.3

t = -0.3 sec

... and...

t = 3 + 3.3

t = 6.3 sec

To understand the two solutions we got, remember that the rocket starts from a raised platform (the roof of a building) and not from the ground. Although we have two x-intercepts, only one of them refers to the time at which the rocket reaches the ground: t_ground = 6.3 sec. The other x-intercept, t = -0.3 sec, is only a theoretical value and doesn't apply to our practical case: if you were to imagine the rocket moving back in time from when it was launched from the roof, its path would intercept the x-axis to the left of the y-axis, which is why we have a negative x-intercept. But, this doesn't happen in practice, so we can eliminate the solution t = -0.3 sec. Our rocket hits the ground 6.3 seconds after being launched.

Answer 2

A model rocket is launched from the roof of a building. It's flight path is modeled by -5t^2 + 30t + 10

where h is the height of the rocket above the ground in meters and t is the time after the launch in seconds.

What is the rocket height after 2 seconds? 

At what time does the rocket reach its maximum height? How far is it above the ground at this time? 

 At what time does the rock et hit the ground?

Ans as h = -5t²+30t+10

at t=0, H = 10, which is height of the building from the ground

at t= 2 sec

h= -5.4+30.2+10

= -20+60 +10 = 50 m

for max height dh/dt = 0

-10t+30 = 0

t=3 sec

Distance from ground = 10 + height after 3 sec

Answer 3

Hello Valeria,

I will explain each part of the question.

h = -5t² + 30t + 10

What is the rocket height after 2 seconds? 

Plug in 2 for t:

h = -5(2)² + 30(2) + 10 = -5(4)+60+10 = -20+60+10 = 50 meters

At what time does the rocket reach its maximum height? How far is it above the ground at this time?

The maximum heigh will be reached when t = -b/2a

b=30, a= -5

t = -30/(-5•2) = 3 seconds

Now, plug in 3 for t to get the maximum height:

h = -5(3)² + 30(3) + 10 = 55 meters

At what time does the rocket hit the ground?

At h=0, the rocket will hit the ground. So, set h=0:

0 = -5t² + 30t + 10

Then, use the quadratic formula to solve:

t = 3 + √11 seconds

Hope this helps!

–Emma