Free Fall Problems: Solving for Speed Without Calculus - Formula Included

In summary, the formula to solve for speed in a free fall problem is v = v_0 + at, where v_0 is the initial velocity (in this case, 0 m/s) and a is the acceleration (in this case, 9.8 m/s^2). Using this formula, the correct answer for the given problem is (c), 5 m/s. This is a simpler approach than using the formula for distance and then working backwards.
  • #1
holly
184
0
I have read through the other free fall problems here on the boards and still cannot correctly solve the following:
One-half second after starting from rest, a freely falling object will have a speed of about: a)10 m/s b)2.5 m/s c)5 m/s d)20 m/s

I find the answer to be (b), 2.5 m/s, but my husband assures me I am not solving it correctly...that I am not using the right formula. I am using a=dt**2 even tho I know acceleration isn't speed...but I don't know what formula to use! This is a physics for health majors class, we don't use calculus. We barely use algebra, and we're all flunking, flunking...

Can some Big Brain please provide me with the formula to solve for speed? The only time I can solve free fall probs is when we are solving for distance and then I use d=1/2gt**2.

Thanking you in advance...
 
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  • #2
You know that the object, whatever it is, has a constant downward acceleration given by g. All you are asked for in this question is the velocity after a given time. The equation you want to use here is

[tex] v = v_0 + at [/tex]

(making c the correct answer).
 
  • #3
Acceleration is defined as "change in speed divided by change in time" of (assumingstarting from rest at t=0) a= v/t so
v= at. That's much simpler that jumping from acceleration to distance and then working back! An object falling freely under gravity, on the surface of the earth, has an acceleration of 9.8 m/s2 downward. After a half second, its speed would be
9.8m/s2(1/2 s)= 4.9 m/s. Much closer to the given 5 m/s than 2.5 m/s.-
 

What is "Yet Another Free Fall Problem"?

"Yet Another Free Fall Problem" is a physics problem that involves an object falling freely under the influence of gravity. It is commonly used to teach students about the laws of motion and the effects of gravity on objects.

What are the key variables in "Yet Another Free Fall Problem"?

The key variables in "Yet Another Free Fall Problem" are the initial height of the object, the acceleration due to gravity, and the time it takes for the object to fall. Other variables, such as air resistance and the mass of the object, may also be considered depending on the specific problem.

How is "Yet Another Free Fall Problem" solved?

There are several equations that can be used to solve "Yet Another Free Fall Problem", including the formula for calculating distance traveled during free fall (d = ½gt²) and the formula for calculating final velocity (v = u + gt). The specific equation used will depend on the given variables and what is being solved for.

What are some real-life applications of "Yet Another Free Fall Problem"?

"Yet Another Free Fall Problem" has many real-life applications, such as calculating the time it takes for an object to fall from a certain height (e.g. a skydiver) or determining the velocity of a falling object (e.g. a ball dropped from a building). It is also used in engineering and construction to calculate the impact of gravity on structures and objects.

What are some common misconceptions about "Yet Another Free Fall Problem"?

One common misconception about "Yet Another Free Fall Problem" is that all objects fall at the same rate, regardless of their mass. In reality, objects with different masses will experience the same acceleration due to gravity, but their final velocities and distances traveled may differ. Another misconception is that air resistance does not affect free fall, when in fact it can significantly alter the results of the problem.

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