how do I solve this?

Question

the acceleration due to gravity on the moon is 5.2ft/sec^2. if an object on the moon is dropped from a height of 4ft, how long is it before it hits the surface of the moon?

Answer 1

Hello, Kate

I think there are two ways you can approach this problem.

1) The Physics Solution

If you've studied Mechanics, you might be familiar with the equation for motion:

X(t) = Xo + Vo*t + 1/2A*t²

Using this equation will allow us to solve for the amount of time it takes for the object to reach the surface of the Moon after being dropped.

First, we have to plug in what we know about the situation.

1. Acceleration on the surface of the Moon is 5.2 ft/sec². A is therefore equal to 5.2. However, we should be careful, and note that gravity is going to pull the object down, and thus reduce its height above the surface, so it will have a negative acceleration. And thus, A = -5.2.
2. The problem says that the object is dropped, and does not mention any initial velocity. So, Vo = 0 ft/s.
3. The object is dropped from an initial height of 4 ft. So, Xo = 4.

With that information, we can rewrite our equation of motion as:

X(t) = 4 + (1/2)*(-5.2)*t² which reduces to X(t) = 4 - 2.6*t²

Now, we're looking for the amount of time it takes for the object to fall, or, in other words, how much time passes between the moment the ball is dropped (t = 0, X(t) = Xo = 4), and the moment that the ball hits the ground (X(t) = 0). So, we have to set our new equation of motion equal to 0.

0 = 4 - 2.6*t² And, now all we have to do is solve for t.

1. Subtract 4 from each side: -4 = -2.6*t²
2. Divide by the coefficient on t (-2.6): 4/2.6 = t² or 1.54 = t²
3. Finally, take the square root of each side to isolate the variable t: 1.24 sec = t

2) The Calculus Solution

This option is identical to the Physics solution, the only difference, however, is that we can't presume to know the equation for motion.

To derive the equation for motion, we have to do some integration.

We're told that the acceleration on the surface of the moon is 5.2ft/sec². Using that, we can set up an equation for acceleration.

A(t) = Ao where Ao = 5.2, but we have to correct for the direction of acceleration, which reduces height. And so, A(t) = -5.2

We can assume that there aren't any additional forces (like air resistance, etc.) that would give us an equation for acceleration that changes with time.

Now, acceleration can be defined as the rate at which an objects velocity changes with time (i.e. the derivative of velocity.) And so, to arrive at an equation of velocity, we just integrate our acceleration equation.

1. V(t) = ∫A(t)dt = ∫(-5.2)dt = -5.2*t +C
2. C here is initial velocity, but we already know Vo is 0. So, V(t) = -5.2*t

And, again, we know that velocity is the rate at which position changes with respect to time, And so, to arrive at our equation of motion, we just need to integrate our equation of velocity.

1. X(t) = ∫V(t)dt = ∫(-5.2*t)dt = (-5.2)*(1/2)*t² + C
2. Which reduces to X(t) = -2.6*t² + C
3. C here is our initial height, Xo, of 4 feet. So, X(t) = -2.6*t² + 4

So, you can see we have arrived at our original equation for motion (just with the terms in a different order), and the steps from here for determining the change in time are identical to the Physics solution.

Hope this helps!

Daniel