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Home Q & A Board Science-Related Homework Help Physics
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FFaraz FFaraz
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10 years ago
This problem is really puzzling me. A projectile P is shot directly upward from the surface of the earth with an initial velocity of 384 ft/s.
1. What is the position equation?
2. What is the average velocity over the interval t=[5, 10]?
3. What is the instantaneous velocity at t=5 and t=10?
4. What is the max height reached?

Please help because I have a calculus exam on this pretty soon. Thanks.
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#1
10 years ago
acceleration = g = -32.2 ft/s^2
 v=velocity ; vo=initial velocity ; s =position
dv/dt =g
int dv = int g dt
v(t) -vo = gt
v(t) =vo+gt
ds/dt = v
int (vo +gt)dt =int ds
s = vo*t +0.5gt^2 (position eqn)
max height v=0
t=vo/g=11.9 s
s(11.9)=2289 ft
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smilez4funsmilez4fun
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10 years ago
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#3
10 years ago
0 - ignore air resistance

1 -  x(T) = 0+ integral of velocity w.r.t. time from t=0 to T
       v(T) = 384 +integral of acceleration w.r.t. time from t=0 to T
      a(T) = constant (about -32 ft/s/s  for x> 0)

3.  when t=5  v(5) = 384 -32*5 =224 ft/s upwards
    v(10) = 384- 320 = 24 ft/s  upwards
   
2. 124 ft/s upwards

4.  max height    reached when v(t) = 0  , t=10.75 seconds
at 384*10.75 - 16*10.75^2
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