An observer 2.5 miles from the launch pad of a space shuttle measures the angle of elevation to the base of the shuttle to be 25 degrees soon after lift off. How high is the shuttle at that instant? (Assume that the shuttle is still moving vertically)
To find the height of the shuttle at that instant, we can use trigonometry.
First, let's label the height of the shuttle as "h" and the distance from the observer to the base of the shuttle as "d".
We are given that the angle of elevation, θ, is 25 degrees and the distance, d, is 2.5 miles.
We can use the tangent function, which is defined as the opposite side (h) divided by the adjacent side (d), to find the height of the shuttle.
Tangent (θ) = Opposite (h) / Adjacent (d)
Tangent (25 degrees) = h / 2.5 miles
Using a calculator, we can find the tangent of 25 degrees to be approximately 0.466.
0.466 = h / 2.5 miles
Now, we can solve for "h" by multiplying both sides of the equation by 2.5 miles:
h = 2.5 miles * 0.466
h ≈ 1.165 miles
Therefore, the height of the shuttle at that instant is approximately 1.165 miles.
To find the height of the shuttle, we can use trigonometry. The observer forms a right triangle with the shuttle and the ground, where the angle of elevation is 25 degrees. The adjacent side of the triangle is the distance between the observer and the base of the shuttle, which is 2.5 miles.
We can use the tangent function to relate the angle of elevation to the height of the shuttle. The tangent of an angle is equal to the opposite side divided by the adjacent side in a right triangle. In this case, the opposite side is the height of the shuttle that we want to find (h).
Therefore, we have:
tan(25 degrees) = h / 2.5 miles
To find h, we can rearrange the equation:
h = 2.5 miles * tan(25 degrees)
Now, we can calculate the height of the shuttle:
h ≈ 2.5 miles * tan(25 degrees)
h ≈ 2.5 miles * 0.4663
So, the height of the shuttle at that instant is approximately:
h ≈ 1.16575 miles
Therefore, the shuttle is approximately 1.16575 miles high.