One way i got mine back was answering questions of others.
Assuming that the stone is thrown vertically... let's say it's a 1 kg stone.It doesn't matter if it's thrown upwards or downwards as (assuming no air friction) it will pass the original throwing point with the same downwards velocity as it had upwards, 3 seconds previously. So it starts with 1/2 m v^2 = 0.5 * 1 * 15^2 = 112.5 J of keThen k.e. gained = gpe lostk.e. gained = m g h = 1 * 10 * 50 = 500 J of Ke gainedso the final (total) ke is 612.5 J which = 1/2 m v^2 = 0.5 v^2 here
so 0.5 v^2 = 612.5so v^2 = 1225so v = 35 m/s
The answer for this problem is clarified through this, the
system is absorbing (+). And now see that it uses that the SURROUNDINGS are
doing 84 KJ of work. Any time a system is overshadowing work done on it by the
surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.
Answer:
Using
Period ( P) is given as
P~√(L/g).
a) since mass has no effect on the period of a pendulum. So, the period will remain 1.8seconds
b) using the formula above ,period varies with the square root of the length. Thus , when the length doubles, the period is multiplied by √2. So, the period is 1.8s*√2 = 2.54s
c) in this case, the period is multiplied by √(1/2).
1.8√(1/2)=1.27s.
d) amplitude of the pendulum doesn't affect the period (unless itsvery high, so, the period is still 1.8s
A) Both players are moving by uniformly accelerated motion, and we can write the position at time t of each of the two players as follows:


where

is the acceleration of the first player

is the acceleration of the second player

is the initial distance between the two players
and where I put a negative sign in front of the acceleration of the second player, since he's moving in the opposite direction of the first player.
The time t at which the two players collide is the time t at which

, therefore:

from whic we find

b) We can use the equation of

to find how far the first player run in t=8.5 s: