A descent vehicle landing on the moon has
a vertical velocity toward the surface of the moon of 26.6 m/s. At the same time, it has a horizontal velocity of 55.9 m/s.
At what speed does the vehicle move along
its descent path?
Answer in units of m/s.
At what angle with the vertical is its path?
Answer in units of degrees.
X = 55.9 m/s.
Y = 26.6 m/s.
a. Speed = sqrt(X^2+Y^2).
b. Tan A = Y/X = 26.6/55.9 = 0.47585.
A = 25.4o S. of E. = 64.6o With vertical
To find the speed of the vehicle along its descent path, we can use the Pythagorean theorem. The total speed is the square root of the sum of the squares of the vertical and horizontal velocities.
Let's calculate the speed:
Speed = √(Vertical velocity^2 + Horizontal velocity^2)
Speed = √(26.6^2 + 55.9^2)
Speed = √(707.56 + 3120.81)
Speed = √3828.37
Speed ≈ 61.92 m/s
So, the vehicle moves along its descent path at a speed of approximately 61.92 m/s.
To find the angle of the path with the vertical, we can use the inverse tangent function. The angle is the arctan of the ratio of the horizontal velocity to the vertical velocity.
Let's calculate the angle:
Angle = arctan(Horizontal velocity / Vertical velocity)
Angle = arctan(55.9 / 26.6)
Angle ≈ arctan(2.102)
Angle ≈ 64.04 degrees
Therefore, the vehicle's path makes an angle of approximately 64.04 degrees with the vertical.
To find the speed of the vehicle along its descent path, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). In this case, the vertical velocity (a) and horizontal velocity (b) are the two sides, and the speed along the descent path (c) is the hypotenuse.
So, the speed along the descent path can be found using the equation:
c^2 = a^2 + b^2
Given that the vertical velocity (a) is 26.6 m/s and the horizontal velocity (b) is 55.9 m/s, we can substitute these values into the equation:
c^2 = (26.6 m/s)^2 + (55.9 m/s)^2
c^2 = 707.56 m^2/s^2 + 3124.81 m^2/s^2
c^2 = 3832.37 m^2/s^2
To find c, we take the square root of both sides:
c = √3832.37 m^2/s^2
c ≈ 61.95 m/s
Therefore, the vehicle moves along its descent path at a speed of approximately 61.95 m/s.
To find the angle (θ) with the vertical, we can use the trigonometric function tangent (tan), which is defined as the ratio of the opposite side (b) to the adjacent side (a) of a right triangle.
tan(θ) = b/a
Substituting the given values:
tan(θ) = 55.9 m/s / 26.6 m/s
θ = tan^(-1)(55.9/26.6)
Using a calculator or trigonometric table, we can find:
θ ≈ 64.7 degrees
Therefore, the angle with the vertical is approximately 64.7 degrees.