Two photographic pictures depict a boy throwing a rock, taken at interval 2 seconds between them? - pictures of boy dirtbike birthday cake
In the second picture of the rock is 10 meters higher than the first. If the stone reaches its maximum height, and what was at that point?
Wednesday, January 27, 2010
Pictures Of Boy Dirtbike Birthday Cake Two Photographic Pictures Depict A Boy Throwing A Rock, Taken At Interval 2 Seconds Between Them?
Subscribe to:
Post Comments (Atom)
7 comments:
We deal with only a vertical motion neglecting air resistance.
a = g = -9.81 m / s ^ 2
at T = 0, y = 0
t = 2, y = 10
y = ut - 1 / 2 GT ^ 2
10 = 2 * u - 2 g
U = 5 + G / S
Maximum height occurs when
/ Dt = 0 u = dy - GT
t = r / g = 1.51 seconds
Maximum = u ^ 2 / (2g) = 11.18 m above the position which described the first image.
Interestingly, the rock is on the way down the second image.
exactly what I had in mind
Certainly not enough information. How do I know what is the maximum height that, unless you know the initial velocity?
There are other variables that come into consideration to find out. You can, I suppose, but not quite right.
In fact, on reflection, simply use the equations of motion!
You have three options:
s [offset] = 10 meters
u [initial rate] Unknown
[To speed] v Unknown
A [acceleration] = -10 m / s ²
t [time] = 2 seconds
[+ Ve Above, So Below-ve]
Connection s = ut + ½ ², u = 5 m / s.
This means that the maximum height of the rock, in a second and a half (v = u + a, where v = 0, and U are the same, which gives t = ½) second.
Using v ² = u ² 2 da than 1.2 meters.
Do not give up, problems are soluble, as a rule to remember, or look at the equations SUVAT, remember that you are only three parameters for these problems.
[I think I have my math is something wrong. But the principle is clear.]
Information is not sufficient
Not enough information.
Can both independent solution
Pictures for the release of the rock.
Post a Comment