Solving Free-Falling Body Homework on Moon Base I

[Edwardo_Elric]

Homework Statement

A lunar lander is making its descent to the moon Base I. THe lander descends slowly under the reto-thrust of its descent engine. The engine is cut off when the lander is 5.0m above the surface and has a downward speed of 1.5m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6m/s^2

Homework Equations

i used
$$V^2 = V_0^2 + 2g(y - y_0)$$

The Attempt at a Solution

required is speed:
$$V = \sqrt{(1.5m/s)^2 + 2(1.6m/s^2)(5.0m)}$$
Speed = 4.27m/s

i don't know about this but what I am confused is that will the gravity be positive when you need a scalar?

 

[PhanthomJay]

Homework Statement

A lunar lander is making its descent to the moon Base I. THe lander descends slowly under the reto-thrust of its descent engine. The engine is cut off when the lander is 5.0m above the surface and has a downward speed of 1.5m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6m/s^2

Homework Equations

i used
$$V^2 = V_0^2 + 2g(y - y_0)$$

The Attempt at a Solution

required is speed:
$$V = \sqrt{(1.5m/s)^2 + 2(1.6m/s^2)(5.0m)}$$
Speed = 4.27m/s

i don't know about this but what I am confused is that will the gravity be positive when you need a scalar?

Velocity, acceleration, and displacement are vector quantities. You chose downward as the positive direction. You could have chosen downward as the negative direction, in which case your equation would have read:
$$V = \sqrt{-1.5m/s)^2 + 2(-1.6m/s^2)(-5.0m)}$$
and you get the same result for V.

 

[Edwardo_Elric]

are u sure?
and thanks a lot
speed is velocity in this problem?

 

[PhanthomJay]

are u sure?
and thanks a lot
speed is velocity in this problem?

No, the velocity is 4.27m/s downward. The speed is the magnitude of the velocity, or 4.27m/s.