A descent vehicle landing on Mars has a vertical velocity toward the surface of Mars of 5.5 m/s. At the same time, it has a horizontal velocity of 2.5 m/s.
(a) At what speed does the vehicle move along its descent path?
(b) At what angle with the vertical is this path?
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The vehicle landing on Mars has a vertical velocity towards the surface of 5.5 m/s. The horizontal velocity of the vehicle is 2.5 m/s.
Adding the two velocities to give the net velocity we get:
The magnitude of the net velocity is sqrt(5.5^2 + 2.5^2) = sqrt(30.25+6.25) = sqrt(36.5) = 6.041 m/s
The angle of the net velocity with the vertical is: arc tan(2.5/5.5) = arc tan(5/11) = 24.44 degrees.
The magnitude of the velocity of the vehicle landing on Mars is 6.041 m/s and its angle with the vertical is 24.44 degrees.
Strictly speaking, this is not a straight forward vector problem because the descent vehicle would have to be slowing from 5.5 m/s to 0 m/s as it landed. Otherwise it would hit the ground at a speed in excess of 5 m/s and suffer significant damage.
But, ignoring that issue, and treating it as a straight vector problem:
Use the Pythagorean Theorem and inverse tangent to solve the problem.
You are going 2.5 m/s horizontal and 5.5 m/s vertical, so in effect you are traveling at an angle downward.
If you construct a right triangle where you start at the origin and go down @ 5.5 m/s and then right @ 2.5 m/s, the hypotenuse of the triangle will give you the actual speed along the descent path.
5.5^2 + 2.5^2 = z^2
z = 6.04 m/s along the descent path.
To find the angle, use inverse tangent.
inverse tangent of (2.5/5.5) = 24.4 degrees.
So your actual descent path is at an angle of 24.4 degrees east of south at a velocity of 6.04 m/s.
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