Complicated Kinematics Problem

[betterfuture]

A baseball is thrown vertically with a speed of 3m/s. Find the total time the ball has been in flight when it has traveled 50cm.

I got stuck halfway. Could someone just assist me on what equations to use and I will do the work myself. Thanks.

---

Members don't see this ad.

 

[Astra]

is it after the ball has traveled 50 cm vertically or horizontally

 

[betterfuture]

IQUOTE="Astra118, post: 17551829, member: 472234"]is it after the ball has traveled 50 cm vertically or horizontally[/QUOTE]

That is what confused me as well. Since he threw the ball vertically, I would assume the question means 50cm vertically. I tried solving it for both as a vertical distance and horizontal distance at different times, and got stuck halfway with weird negative numbers.

 

[To be MD]

A baseball is thrown vertically with a speed of 3m/s. Find the total time the ball has been in flight when it has traveled 50cm.

I got stuck halfway. Could someone just assist me on what equations to use and I will do the work myself. Thanks.

Hey, I think either you meant to type 30 m/s for initial velocity or the problem has an error the way it is written, and it should be 30 m/s. An object can't get 50 m of total distance traveled with a initial velocity of 3 m/s on planet earth in any direction.

If you throw something straight up at 30 m/s, the velocity is only in the vertical direction. That means that you can use vfinal^2 = vinitial^2 + 2*a*d

This equation is helpful because if you throw something straight up, at its maximum height, the velocity of the object is 0 m/s. Plugging that in with an acceleration due to gravity of ~10 m/s^2, we can solve for the height.

--> 0 = 30^2 + 2(-10)d ---> -900 = -20 d ---> d = 90/2 or 45 meters. That gives us the height at the peak of the arc. **Remember, the question is not talking about displacement, it's asking about time after a distance is traveled. **

_______

Now, we know the question is asking find time when the ball has traveled a total of 50m, and we're pretty much there because we can solve for time when it has traveled 45 meters! We can use the definition of acceleration to solve for that. a = ∆v / t .... We know a as we defined it is - 10 m/s^2 and that equals to change in velocity (v final - v initial ... or 0 - 30 m/s) / t

--> -10 = -30/t --> t = 30/10 --> t = 3

On an MCAT exam, you'd stop right there. The answer choice is going to be something a little greater than t = 3 seconds.

____

If you want an exact answer, we know that the acceleration is still -10 m/s^2 even when the object has 0 velocity at its peak height, and now we can treat it like we're on a tower that's 45 meters tall and we drop the object off the side.

Now you're going to use, d = vinitial * (t) + 1/2 at^2 .... The distance left we have to solve for is -5 meters (45 to get to the top, 5 more to get a total of 50 meters traveled)... Plug in your numbers to solve for the other time:

-5 = 0 * t + 1/2 (-10)*t^2 ---> -5 = 1/2 (-10) t^2 ---> t^2 = 1 --> t = 1 s
**notice, since we've defined falling downwards as negative both acceleration and 5 are negative.
____

Final answer, at 50 meters total distance traveled, the object has been in flight for (3 + 1 = 4) 4 seconds.

 

[theonlytycrane]

OP said 50 cm, not 50 m.

Use v_y = 3 m/s (vertical), distance = .5 m, a_g = -10 m/s^2 and think of throwing a baseball straight up.

Use y = y_initial + v_y * t + 1/2 * a * t^2. Where y_initial = 0 (assume you throw the baseball up from ground level).

 

[betterfuture]

@To be MD Thank you very much for you generous help. I noticed the numbers were different but solving the problem using any numbers would essentially use the same steps/methods. The steps seemed a bit lengthy, but they obviously gave us the answer.

OP said 50 cm, not 50 m.

Use v_y = 3 m/s (vertical), distance = .5 m, a_g = -10 m/s^2 and think of throwing a baseball straight up.

Use y = y_initial + v_y * t + 1/2 * a * t^2. Where y_initial = 0 (assume you throw the baseball up from ground level).

I used that equation. And I got stuck there.

I got 2(o.5)/-10 = 3t + t^2. How do you get t by itself on the right hand side? I know maybe my algebra skills are probably a little weak right now, since its been a couple of years since I last took any algebra. Any help here?

 

[theonlytycrane]

bring all the terms to one side and use the quadratic formula to solve for solutions for t!