A model rocket is launched from rest with an upward acceleration of 6.30 m/s2 and, due to a strong wind, a horizontal acceleration of 1.50 m/s2.

How far is the rocket from the launch pad 5.30 s later when the rocket engine runs out of fuel?

VERTICAL PROBLEM:

y = (1/2) 6.3 t^2
= 3.15 (5.3)^2

HORIZONTAL PROBLEM:
x = (1/2)1.5 t^2
= .75 (5.3)^2

d = sqrt(x^2+y^2)

Well, if the rocket engine runs out of fuel, I guess it's no longer fuel for thought! But let's calculate the rocket's distance from the launch pad anyway.

To find the rocket's distance, we need to find the horizontal and vertical displacements separately.

First, let's find the horizontal displacement (d_horizontal):
Using the formula: d_horizontal = (initial velocity) x (time) + (0.5) x (horizontal acceleration) x (time)^2

Since the rocket starts from rest, the initial velocity is zero. So, d_horizontal = (0.5) x (1.50 m/s^2) x (5.30 s)^2

Now, let's find the vertical displacement (d_vertical):
Using the formula: d_vertical = (initial velocity) x (time) + (0.5) x (vertical acceleration) x (time)^2

Since the rocket starts from rest vertically, the initial velocity is also zero. So, d_vertical = (0.5) x (6.30 m/s^2) x (5.30 s)^2

Finally, we can find the total distance using the Pythagorean theorem:
Total distance = square root of [(d_horizontal)^2 + (d_vertical)^2]

But hey, before we do all these calculations, let me tell you, this rocket is really going places! So here we go, calculate away!

To find the distance the rocket is from the launch pad 5.30 seconds later when the rocket engine runs out of fuel, we need to calculate the vertical and horizontal displacements separately and then use the Pythagorean theorem to find the total distance.

First, let's calculate the vertical displacement:

Given: Initial velocity (u) = 0 m/s, acceleration (a) = 6.30 m/s^2, time (t) = 5.30 s

The formula for vertical displacement is given by:
s = ut + 0.5at^2

Substituting the given values, we get:
s = 0 * 5.30 + 0.5 * 6.30 * (5.30)^2

Calculating this gives us:
s = 0 + 0.5 * 6.30 * 28.09
s = 0 + 88.7205
s = 88.7205 m

Next, let's calculate the horizontal displacement:

Given: Initial velocity (u) = 0 m/s, acceleration (a) = 1.50 m/s^2, time (t) = 5.30 s

The formula for horizontal displacement is given by:
s = ut + 0.5at^2

Substituting the given values, we get:
s = 0 * 5.30 + 0.5 * 1.50 * (5.30)^2

Calculating this gives us:
s = 0 + 0.5 * 1.50 * 28.09
s = 0 + 21.135
s = 21.135 m

Now, let's use the Pythagorean theorem to find the total distance from the launch pad:
d = √(vertical displacement^2 + horizontal displacement^2)

Substituting the calculated values, we get:
d = √(88.7205^2 + 21.135^2)

Calculating this gives us:
d = √(7873.303 + 445.138)
d = √(8318.441)
d = 91.13 m (rounded to two decimal places)

Therefore, the rocket is approximately 91.13 meters away from the launch pad 5.30 seconds later when the rocket engine runs out of fuel.

To find the distance the rocket is from the launch pad when the rocket engine runs out of fuel, we can break down the motion of the rocket into its vertical and horizontal components.

First, let's analyze the vertical motion of the rocket. We are given that the rocket has an upward acceleration of 6.30 m/s². We can use the equation of motion, \(s = ut + \frac{1}{2}at^2\), where "s" is the vertical displacement, "u" is the initial vertical velocity, "a" is the vertical acceleration, and "t" is the time.

Since the rocket is launched from rest, the initial vertical velocity (u) is 0 m/s. Therefore, the equation simplifies to \(s = \frac{1}{2}at^2\).

Plugging in the given values, we have \(s = \frac{1}{2} \times 6.30 \, \text{m/s²} \times (5.30 \, \text{s})^2\).
Calculating this expression will give us the vertical displacement of the rocket from the launch pad.

Next, let's analyze the horizontal motion of the rocket. We are given that the rocket experiences a horizontal acceleration of 1.50 m/s² due to the wind. However, since there is no initial horizontal velocity mentioned, we can assume that the rocket starts from rest horizontally as well.

To find the horizontal displacement, we can use the equation \(s = ut + \frac{1}{2}at^2\) again, where "s" is the horizontal displacement, "u" is the initial horizontal velocity, "a" is the horizontal acceleration, and "t" is the time.

Since the rocket starts from rest, the initial horizontal velocity (u) is 0 m/s. Therefore, the equation simplifies to \(s = \frac{1}{2}at^2\).

Plugging in the given values, we have \(s = \frac{1}{2} \times 1.50 \, \text{m/s²} \times (5.30 \, \text{s})^2\).
Calculating this expression will give us the horizontal displacement of the rocket from the launch pad.

Finally, to find the distance the rocket is from the launch pad, we can use the Pythagorean theorem. The distance will be the square root of the sum of the squares of the vertical and horizontal displacements.

In equation form, the distance (d) is given by \(d = \sqrt{s_{\text{vertical}}^2 + s_{\text{horizontal}}^2}\).

Now, plug in the values we calculated for the vertical and horizontal displacements and evaluate the expression to find the distance of the rocket from the launch pad when the rocket engine runs out of fuel.