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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.

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

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.
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.

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.

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.

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:)