From what I can see, part (a) is right. The first thing that would be good to do is, to split the acceleration and the deceleration sections up. As you know the velocity of the first part, you need to find the velocity of the second part. How would you go about doing that, as you have the time...
I checked your answers and everything, and from what I know, you are completely correct.
By the way, as advice, if you have not already done it in other examples it is good show working on the site, and if possible a vector diagram.
you are on track with the answer, but why use 11ms-1? (as you stated earlier that it was the terminal velocity, and you were going to use half of the terminal velocity.
Also, the equation does not need to be so complicated. Look at these ones, and figure out which one is needed, by knowing the...
Read up on anything you might be able to find on http://www.regentsprep.org/Regents/physics/phys06/bcentrif/default.htm.
P.S. To the senior members: was this the right thing to do; provide a link?
How did you get the to answer of 3.9 seconds and 37.6 ms-[sup]1[sup]? From what you have given me in section one (the known data) I can only find the horizontal distance and the launch angle, and I cannot understand how you find your given answer from that. Is there a missing piece of data that...
That very well might work, the only reason why it might not, is because the camera might not be able to record enough frames in a second, for you to be able to break the recording down and analyse it properly. The strob light might help it do that, but I cannot say for sure, as I have never...
Okay, Make a vector diagram of the ramp, the initial velocity at 30 degrees and you can then find the vertical velocity.
Remember this: when trying to work out a problem that at first does not make sense, draw a vector diagram (if applicable), it is remarkable how useful they really are.
I assume it means that the baseball landed at to the horizontal 28 degrees. Also, know that the 7.5 m is how much higher the ball landed at the end will make it a lot harder.
What do you state gravity as?
the gravitational potential energy equation is this:
U_g=mgh
"U" is the energy in joules, "m" is mass in kg, "g" is the gravity value in ms-2, and "h" is the maximum height of the object in metres. the value of "g" is normally taken as 9.8 or 10ms-12...
When you are substituting the velocity into the equation, the left side of the equation. The equation normally should look more like this:
m_a*u_a-m_b*u_b=m_a*v_a+m_b*v_b
By the way, I state "u" as the initial velocity of the object(s), and "v" as the final.
Yeah, as said, I think the question needs to be clarified somewhat, as I am slightly confused to how you came to get the final velocity and its' components.
When the baseball is hit at \theta degrees to the horizontal, does the baseball reach it's maximum height (7.5 m) and fly down again?
I am also a bit puzzled about what you mean the 28 degrees is.
:blushing:
No, no, it is me. There are other equations:
s=\frac{1}{2}(u+v)t
v=u+at
s=ut+\frac{1}{2}at^2
v^2=u^2+2as
Which one is the most appropriate? Also, "v" is the final and "u" is the initial velocity. from knowing the answer to the equation you use, you then just need to do a...