Posted by Xiang ! on Monday, August 26, 2013 at 8:45am.
first stone
h=1/2 g t^2
second stone:
h=1/2 g (t-2)^2
set the equations the same, solve for time t it took the first stone
solve for height of build, h.
for speed, v^2=2gh
2. you are given h, solve for t
h=1/2 g t^2
speed is a combination of horizontal and vertical speed.
vertical=sqrt(2gh)
horizontal given
speed=sqrt(horizontal^2+vertical^2)
distance=horizontal speed*time t
can u help me solve this also ?
A car from rest and travels for 5.0 s with uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform deceleration of -2.0 m/s2. If the brakes are applied for 3.0 s.
i) How fast is the car going at the end of the braking period?
ii) How far has it gone?
iii) Sketch a graph of velocity versus time for the motion.
Hmmm.
1) If y is the stone's height at t, and h is the building height, I get
first stone: y = h - 4.9t^2
2nd stone: y = h - 25(t-2) - 4.9(t-2)^2
i: t = 5.63
the rest is easy...
For the car problem, just use your standard equations of motion.
accelerating: v = at = (1.5)(5) = 7.5
decelerating: v = 7.5 - 2(3) = 1.5
distance: 1/2 (1.5)(25) + 7.5(3) + 1/2 (-2)(9) = 32.25
1a. d1 = 0.5g*t^2 = 4.9*2^2 = 19.6 m. =
Distance fallen by the 1st stone after 2 s.
V1^2 = Vo1^2 + 2g*d
V1^2 = 0 + 19.6*19.6 = 384.16
V1 = 19.6 m/s = Velocity of 1st stone after 2 s.
d1 = d2-19.6 m.
Vo1*t + 0.5g*t^2 = Vo2*t+0.5t^2-19.6
Vo1*t-Vo2*t+4.9t^2-4.9t^2 = -19.6
19.6t-25t = -19.6
-5.4t = -19.6
Tf2 = 3.63 s. = Fall time of 2nd stone.
Tf1 = 3.63 + 2 = 5.63 s. = Fall time of
1st stone.
1b. h = Vo*t + 0.5g*t^2
h = 0 + 4.9(5.63)^2 = 155.3 m.
1c. V1^2 = Vo^2 + 2g*h
V1^2 = 0 + 19.6*155.3 = 3,043.88
V1 = 55.2 m/s.
V2^2 = 25^2 + 19.6*155.3 = 3668.88
V2 = 60.6 m/s.