A descent vehicle landing on the moon has
a vertical velocity toward the surface of the
moon of 36.8 m/s. At the same time, it has a
horizontal velocity of 52.6 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 ◦
.
X = 52.6 m/s.
Y = 36.8 m/s.
Speed = sqrt(X^2+Y^2).
Tan A = Y/X = 36.8/52.6 = 0.69962
A = 34.98o CCW = 55.02o From vertical.
1. 64.195
To find the speed of the vehicle along its descent path, we can use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle (in this case, the total velocity of the vehicle) is equal to the sum of the squares of the other two sides (in this case, the vertical and horizontal velocities).
The vertical velocity towards the surface of the moon is 36.8 m/s, and the horizontal velocity is 52.6 m/s.
Applying the Pythagorean theorem:
Total velocity^2 = vertical velocity^2 + horizontal velocity^2
Total velocity^2 = (36.8)^2 + (52.6)^2
Total velocity^2 = 1353.44 + 2760.76
Total velocity^2 = 4114.2
Taking the square root of both sides to find the total velocity:
Total velocity = √4114.2
Total velocity ≈ 64.12 m/s
Therefore, the speed of the vehicle along its descent path is approximately 64.12 m/s.
To find the angle with the vertical, we can use the inverse tangent function. The angleθ is given by:
θ = arctan(horizontal velocity / vertical velocity)
θ = arctan(52.6 / 36.8)
θ ≈ 53.52°
Therefore, the angle with the vertical is approximately 53.52°.
To find the speed at which the vehicle moves along its descent path, we can use the concept of vector addition. The speed along the descent path is equal to the magnitude of the resultant velocity vector.
We can find the resultant velocity vector by using the Pythagorean theorem to combine the vertical and horizontal velocities of the vehicle.
Using the given information:
Vertical velocity, vy = 36.8 m/s
Horizontal velocity, vx = 52.6 m/s
The resultant velocity vector is given by:
Resultant velocity, V = sqrt(vx^2 + vy^2)
Plugging in the values:
V = sqrt((52.6 m/s)^2 + (36.8 m/s)^2)
Calculating:
V ≈ sqrt(2760.76 m^2/s^2 + 1354.24 m^2/s^2)
V ≈ sqrt(4115 m^2/s^2)
Therefore, the speed at which the vehicle moves along its descent path is approximately 64.22 m/s.
To find the angle with the vertical, we can use trigonometry. The angle can be determined by calculating the arc tangent of the vertical velocity divided by the horizontal velocity.
Using the given information:
vy = 36.8 m/s
vx = 52.6 m/s
The angle, θ, can be found by:
θ = atan(vy/vx)
Plugging in the values:
θ = atan(36.8 m/s / 52.6 m/s)
Calculating:
θ ≈ atan(0.701) ≈ 34.96°
Therefore, the angle with the vertical is approximately 34.96°.