A descent vehicle landing on the moon has
a vertical velocity toward the surface of the moon of 35.4 m/s. At the same time, it has a horizontal velocity of 57.2 m/s.
At what speed does the vehicle move along
its descent path?
At what angle with the vertical is its path?
1. V = sqrt((57.2)^2 + (35.4)^2)=67.3m/s.
2. tanA = -35.4 / 57.2 = -0,6189
A = -31.75 deg.,CW.
A = -31.75 + 360 = 328 deg.,CCW.
A = 328 - 270 = 58 deg. with vertical.
To determine the speed at which the vehicle moves along its descent path, we can use the concept of vector addition. The resultant velocity is the combination of the vertical and horizontal velocities.
To calculate the speed, we need to find the magnitude of the resultant velocity. We can use the Pythagorean theorem, which states that for a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.
Let's call the vertical velocity Vv and the horizontal velocity Vh.
Vv = 35.4 m/s
Vh = 57.2 m/s
Using the Pythagorean theorem:
Resultant velocity (V) = sqrt(Vv^2 + Vh^2)
V = sqrt((35.4 m/s)^2 + (57.2 m/s)^2)
V = sqrt(1253.16 m^2/s^2 + 3276.64 m^2/s^2)
V = sqrt(4529.8 m^2/s^2)
V ≈ 67.34 m/s
Therefore, the speed at which the vehicle moves along its descent path is approximately 67.34 m/s.
Next, let's determine the angle with the vertical. We can use trigonometry to find the angle.
Let's call this angle θ.
θ = arctan(Vh / Vv)
θ = arctan(57.2 m/s / 35.4 m/s)
θ = arctan(1.615)
θ ≈ 58.52°
Therefore, the angle with the vertical is approximately 58.52°.