kinematics (urgent)

1. The time that the pot passes through the window is 0.5/2 s = 0.25 s
Apply equation s = ut + (1/2)at^2 for the upward journey
with s = 2m, t = 0.25 s, a =-g(=-9.81 m/s2), u =?
hence, 2 = 0.25u + (1/2)(-9.81)(0.25^2)
u = 6.77 m/s

For the journey from the bottom of the window to the highest point, apply the equation" v^2 = u^2 + 2a.s
with v = 0 m/s, u = 6.77 m/s, a = -g, s =?
hence, 0 = 6.77^2 + 2.(-9.81)s
s = 2.339 m
Therefore, the pot reaches (2.339 - 2) m = 0.339 m above the window top.
[your given answer is the height from the window bottom]

2. For the journet passing through the window, let U be the velocity at the window bottom, use equation: s = vt-(1/2)gt^2
with s = -1.2 m, v = U, g = -9.81 m/s2, t = 0.125 s
hence, -1.2 = 0.125U - (1/2).(-9.81).(0.125^2)
U = -10.213 m/s

For the jourey from the window bottom to the ground, apply equation:v = u +at
with u = -10.213 m/s, a = -9.81 m/s2, t = 1 s, v =?
v = [-10.213 + (-9.81) x 1] m/s = -20.023 m/s

For the journey from the roof to ground, use equation: v^2 = u^2 + 2a.s
with v = -20.023 m/s, u = 0 m/s, a = -9.81 m/s2, s =?
(-20.023)^2 = 2.(-9.81).s
s = -20.43 m
The height of the building is 20.43 m

3. The difference in speed is the difference in areas bounded by the two curves with the time axis.
You need to calculate the areas bit by bit.

I think "differential" is a mathematical term, not a term in physics.









2011-10-01 23:10:34 補充：
sorry...I made a wrong calculation in Q(1). u should be 9.225 m/s, this gives the height reached is 2.34m from top of window.