NASA Launches A Rocket At T = 0 T=0 T = 0 Seconds. Its Height, In Meters Above Sea Level, As A Function Of Time Is Given By H ( T ) = − 4.9 T 2 + 70 T + 269 H(t)=-4.9 T^2+70 T+269 H ( T ) = − 4.9 T 2 + 70 T + 269 . For Each Of The Answers Below, Put Your Numerical Answer, Rounded To 1 Decimal Place, In The

Introduction

In this article, we will explore the height of a rocket launched by NASA as a function of time. The height of the rocket is given by the quadratic equation $h(t)=-4.9 t^2+70 t+269$, where $t$ is the time in seconds and $h(t)$ is the height in meters above sea level. We will calculate the height and velocity of the rocket at different times and analyze the results.

Calculating Height

To calculate the height of the rocket at a given time, we can plug the value of $t$ into the equation $h(t)=-4.9 t^2+70 t+269$. Let's calculate the height at $t=0$, $t=1$, $t=2$, and $t=3$ seconds.

Height at $t=0$ seconds

At $t=0$ seconds, the height of the rocket is given by:

$h(0)=-4.9 (0)^2+70 (0)+269$

$h(0)=269$

The height of the rocket at $t=0$ seconds is 269 meters.

Height at $t=1$ second

At $t=1$ second, the height of the rocket is given by:

$h(1)=-4.9 (1)^2+70 (1)+269$

$h(1)=-4.9+70+269$

$h(1)=334.1$

The height of the rocket at $t=1$ second is 334.1 meters.

Height at $t=2$ seconds

At $t=2$ seconds, the height of the rocket is given by:

$h(2)=-4.9 (2)^2+70 (2)+269$

$h(2)=-19.6+140+269$

$h(2)=389.4$

The height of the rocket at $t=2$ seconds is 389.4 meters.

Height at $t=3$ seconds

At $t=3$ seconds, the height of the rocket is given by:

$h(3)=-4.9 (3)^2+70 (3)+269$

$h(3)=-44.1+210+269$

$h(3)=435.9$

The height of the rocket at $t=3$ seconds is 435.9 meters.

Calculating Velocity

To calculate the velocity of the rocket at a given time, we need to find the derivative of the height function $h(t)=-4.9 t^2+70 t+269$. The derivative of $h(t)$ is given by:

$h'(t)=-9.8 t+70$

Let's calculate the velocity at $t=0$, $t=1$, $t=2$, and $t=3$ seconds.

Velocity at $t=0$ seconds

At $t=0$ seconds, the velocity of the rocket is given by:

$h'(0)=-9.8 (0)+70$

$h'(0)=70$

The velocity of the rocket at $t=0$ seconds is 70 m/s.

Velocity at $t=1$ second

At $t=1$ second, the velocity of the rocket is given by:

$h'(1)=-9.8 (1)+70$

$h'(1)=-9.8+70$

$h'(1)=60.2$

The velocity of the rocket at $t=1$ second is 60.2 m/s.

Velocity at $t=2$ seconds

At $t=2$ seconds, the velocity of the rocket is given by:

$h'(2)=-9.8 (2)+70$

$h'(2)=-19.6+70$

$h'(2)=50.4$

The velocity of the rocket at $t=2$ seconds is 50.4 m/s.

Velocity at $t=3$ seconds

At $t=3$ seconds, the velocity of the rocket is given by:

$h'(3)=-9.8 (3)+70$

$h'(3)=-29.4+70$

$h'(3)=40.6$

The velocity of the rocket at $t=3$ seconds is 40.6 m/s.

Discussion

Time (s)	Height (m)	Velocity (m/s)
0	269.0	70.0
1	334.1	60.2
2	389.4	50.4
3	435.9	40.6

The height of the rocket increases as time increases, while the velocity of the rocket decreases as time increases. This is because the rocket is accelerating downward due to gravity, causing its velocity to decrease over time.

Conclusion

Introduction

In our previous article, we explored the height of a rocket launched by NASA as a function of time. The height of the rocket is given by the quadratic equation $h(t)=-4.9 t^2+70 t+269$, where $t$ is the time in seconds and $h(t)$ is the height in meters above sea level. In this article, we will answer some frequently asked questions about the rocket launch.

Q: What is the initial height of the rocket?

A: The initial height of the rocket is 269 meters above sea level.

Q: What is the maximum height of the rocket?

A: To find the maximum height of the rocket, we need to find the vertex of the parabola given by the equation $h(t)=-4.9 t^2+70 t+269$. The x-coordinate of the vertex is given by:

$t=-\frac{b}{2a}$

where $a=-4.9$ and $b=70$. Plugging in these values, we get:

$t=-\frac{70}{2(-4.9)}$

$t=7.14$

The maximum height of the rocket is given by plugging this value of $t$ into the equation $h(t)=-4.9 t^2+70 t+269$:

$h(7.14)=-4.9 (7.14)^2+70 (7.14)+269$

$h(7.14)=435.9$

The maximum height of the rocket is 435.9 meters above sea level.

Q: What is the velocity of the rocket at the maximum height?

A: To find the velocity of the rocket at the maximum height, we need to find the derivative of the height function $h(t)=-4.9 t^2+70 t+269$. The derivative of $h(t)$ is given by:

$h'(t)=-9.8 t+70$

Plugging in the value of $t$ that gives the maximum height, we get:

$h'(7.14)=-9.8 (7.14)+70$

$h'(7.14)=-40.6+70$

$h'(7.14)=29.4$

The velocity of the rocket at the maximum height is 29.4 m/s.

Q: How long does it take for the rocket to reach the maximum height?

A: To find the time it takes for the rocket to reach the maximum height, we need to find the value of $t$ that gives the maximum height. We already found this value to be $t=7.14$ seconds.

Q: What is the acceleration of the rocket?

A: The acceleration of the rocket is given by the second derivative of the height function $h(t)=-4.9 t^2+70 t+269$. The second derivative of $h(t)$ is given by:

$h''(t)=-9.8$

The acceleration of the rocket is -9.8 m/s^2.

Q: What is the terminal velocity of the rocket?

A: The terminal velocity of the rocket is the velocity at which the acceleration of the rocket is zero. Since the acceleration of the rocket is given by $h''(t)=-9.8$, the terminal velocity is given by:

$v=-\frac{h''(t)}{g}$

where $g=9.8$ m/s^2. Plugging in the value of $h''(t)$, we get:

$v=-\frac{-9.8}{9.8}$

$v=1$

The terminal velocity of the rocket is 1 m/s.

Conclusion

In this article, we answered some frequently asked questions about the rocket launch. We found that the initial height of the rocket is 269 meters above sea level, the maximum height is 435.9 meters above sea level, and the velocity at the maximum height is 29.4 m/s. We also found that the time it takes for the rocket to reach the maximum height is 7.14 seconds, the acceleration of the rocket is -9.8 m/s^2, and the terminal velocity of the rocket is 1 m/s.