Monday

October 20, 2014

October 20, 2014

Posted by **Xiang !** on Monday, August 26, 2013 at 8:45am.

i) How long did it take the first stone to reach the ground?

ii) How high is the building?

iii) What are the speeds of the two stones just before they hit the ground?

2)A jet fighter is traveling horizontally with a speed of 111 m/s at an altitude of 3.00×102 m, when the pilot accidentally releases an outboard fuel tank.

i) How much time elapses before the tank hits the ground

ii) What is the speed of the tank just before it hits the ground

iii) What is the horizontal distance traveled by the tank.

- Physics !! -
**bobpursley**, Monday, August 26, 2013 at 10:05amfirst 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

- Physics !! -
**Xiang !**, Monday, August 26, 2013 at 10:46amcan 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.

- Physics !! -
**Steve**, Monday, August 26, 2013 at 12:40pmHmmm.

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...

- Physics !! -
**Steve**, Monday, August 26, 2013 at 1:10pmFor 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

- Physics !! -
**Henry**, Monday, August 26, 2013 at 8:56pm1a. 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.

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