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.

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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Scottsdale, AZ

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 = v_{0}sin30 - gt = 0

t = (1/2)350/9.8 = 17.9 sec (Here g = 9.8 m/s^2, v_{0} = 350 m/s)

So, by symmetry the total flight time = 2*17.9 = 35.7 sec.

2) v^2 = (v_{0}sin30)^2 - 2gh

When v = 0, h = hmax.

So, the maximum height = (v_{0}sin30)^2 / (2g) = 1562.5 m

3) R = v_{0}cos30 * (2t) = 350cos30 * 35.7 = 10821. m

Willowbrook, IL

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=v_{0}cos(30)*t;

b) Vertical:

y=v_{0}sin(30)*t-gt^{2}/2; minus sign here shows that g is directed downward.

Equate y to zero and solve for t:

v_{0}sin(30)*t-gt^{2}/2=0; t=0 or t=2v_{0}sin(30)/g;

t=0 correspond to initial moment, t=2v_{0}sin(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/s^{2} 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*17.5^{2}/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=v_{0}cos(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.

New Wilmington, PA

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= v_{ix}t , Δy=-½gt²+v_{iy}t

Set the second of these equal to zero and you can divide everything by t (since t≠0):

-½gt+v_{iy} = 0

Therefore,

t = 2v_{iy}/g

We need to calculate the components of the initial velocity:

v_{ix} = 350 cos(30) = 303 m/s

v_{iy} = 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= v_{ix}t= 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.

Blacksburg, VA

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:)

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## Comments

_{0}sin30 - gt = 0 tells you this is in the direction parallel to g.