1

# 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.

### 4 Answers by Expert Tutors

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
2
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.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
1
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.

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
1
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.