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Find its time of flight?

A shell was fired from ground level with an initial speed of 350 m/s at an angle of 30 degrees to the horizontal. Assuming there is no air resistance, calculate:
 
a) its time of flight;
b) its maximum altitude;
c) its range.

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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
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Projectile motion has symmetry. If you use the symmetry, you can get the answer lot faster.
1) At the top, the velocity is zero.
v = v0sin30 - gt = 0
t = (1/2)350/9.8 = 17.9 sec (Here g = 9.8 m/s^2, v0 = 350 m/s)
So, by symmetry the total flight time = 2*17.9 = 35.7 sec.
 
2) v^2 = (v0sin30)^2 - 2gh
When v = 0, h = hmax.
So, the maximum height = (v0sin30)^2 / (2g) = 1562.5 m
 
3) R = v0cos30 * (2t) = 350cos30 * 35.7 = 10821. m

Comments

1. The projectile's velocity is not zero at the top. It is moving at 350 m/s at 0°.
 
2. Symmetry arguments can only be used if the projectile lands at the same level it started, which was not specified in this problem.
It's 350cos(30) m/s on top, sorry.
1) v = v0sin30 - gt = 0 tells you this is in the direction parallel to g.
2) R is calculated based on ending at ground level here.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
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1) Vertical direction is up, horizontal is towards the firing direction, origin is at the starting point.
 
2) Write two equations of motion:
   a) Horizontal:
      x=v0cos(30)*t;
  b) Vertical:
     y=v0sin(30)*t-gt2/2; minus sign here shows that g is directed downward.
 
Equate y to zero and solve for t:
 
v0sin(30)*t-gt2/2=0; t=0 or t=2v0sin(30)/g;
t=0 correspond to initial moment, t=2v0sin(30)/g correspond to the final moment, when the shell hits the ground. This is the time of flight. Calculation yields t=35 s (g=10 m/s2 was used).
 
The time of flying up to the maximum altitude equals to the time of descent, so it took 35/2=17.5 s for the shell to reach maximum height. Plug in t=17.5 into the vertical equation of motion to get:
 
h=350*sin(30)*17.5-10*17.52/2≈1531 m; This is the maximum altitude.
 
The range can be found from the horizontal equation of motion by plugging in the total time of flight, 35 s, into the equation.
 
L=v0cos(30)*t=350*cos(30)*35≈10609 m=10.609 km.
 
Answers:
 
a) t=35 s
b) H=1531 m.
c) L=10609 m.
 
It is worth mentioning that in reality the range will be just a few kilometers for such a shell, due to air resistance.
 
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
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I'm assuming that the shell lands back at ground level, or else they would have to tell us how far above/below ground level it lands. In this case, Δy=0.
 
We use the kinematic equations
 
Δx= vixt ,  Δy=-½gt²+viyt
 
Set the second of these equal to zero and you can divide everything by t (since t≠0):
 
-½gt+viy = 0
 
Therefore,
 
t = 2viy/g
 
We need to calculate the components of the initial velocity:
 
vix = 350 cos(30) = 303 m/s
viy = 350 sin(30) = 175 m/s
 
Therefore,
 
t = 2(175)/9.8 = 35.7 s
 
is the flight time.
 
The range follows from the other equation,
 
Δx= vixt= 303 (35.7) =10 800 m.
 
The maximum altitude is the height at half the flight time, t= 35.7/2 = 17.9 s.
 
At this point,
 
Δy=-½(9.8)(17.9)²+175(17.9) = 1560 m.
 
 
 
Brad M. | Professional Tutor: STEM plus Business, Accounting, and InvestmentProfessional Tutor: STEM plus Business,...
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Sun -- reasoning ... sin 30 ~ 30/60 = 1/2 ... Vup 175m/s takes 17.5s of uptime at 10m/s/s ... ave Vup is 90m/s acting 17.5s makes max ht of 1750-175 or 1575m ..  airtime is 35s ... range is 85% of 350*35 or about ~30*350 ==> range of 10.5km  Best wishes, sir:)