(Solved) How do you find the launch angle and flight duration?

smiley13

wrote...

Go to Answer

Posts: 21

Rep:

11 years ago

How do you find the launch angle and flight duration?

If a projectile is launched at 100 m/s and flies 1000m before detonating 10m off the ground, what was the launch angle and the time of the flight?

Read 239 times

3 Replies

Report

Replies

Answer accepted by topic starter

tonyagodsey

wrote...

Posts: 38

Rep:

11 years ago

Report

Related Topics

Projectile Motion: How do I find velocity when only given the angle of launch and the distance traveled?
Solved If you plot range of projectile against the angle of launch, what will the slope give you?
In triangle ABC, angle C is a right angle. Find the measure of side a if the mea
Solved In triangle ABC, angle C is a right angle. Find the measure of side a if the mea
Solved If lift and duration remain constant and the lobe center angle decreases, then ________
Solved Anna Templeton is preparing to launch a home security firm. The team of people that will launch Anna
Solved Elizabeth fell down a flight of stairs. The angle between her right foot and the tibia increased as ...
Pulsing refers to the grouping of advertisements in flight, at different weights and duration over ...

n_m_m

wrote...

11 years ago

As with many trajectory problems, there are 2 launch angles which will give the same range but different times. The launch angles are the roots of:

[gx²/2v²]*tan²? - x*tan? + gx²/2v² + y = 0 = 490*tan²? - 1000*tan? + 500

tan? = 1.1647, 0.8761
?1 = 49.351°
?2 = 41.22°
Hmax1 = Vy1²/2g = (vsin?1)²/2g = 293.7 m; let h1 = Hmax1 - 10 = 283.7 m
Hmax2 = Vyi2²/2g = 221.54 m; let h2 = Hmax2 - 10 = 211.54 m
The Hmax values are the rising height; h is the falling height.

t1 = Vyi²/g + ?[2h1/g] = Visin?1/g + ?[2*283.7/g] = 100sin?1/g + ?[2*283.7/g] = 15.351 sec
t2 = 100sin?2/g + ?[2*h2/g] = 13.294 sec

Report

darkshinra

wrote...

11 years ago

d_horizontal = vi*cos(theta)*t [There is no acceleration]
d_vertical = vi*sin(theta)*t - 1/2 *a * t^2

d_horizontal = 1000 m
d_vertical = 10 m
vi = 100 m/s
a = 9.81
t = Neutral Face

theta =

Two equations, two unknowns. We should be able to get it although I can tell you that it is not going to be easy.

Solve for t in d_horizontal.

1000 / (100 * cos(theta)) = t
t = 10/(cos(theta))

10 = 100*sin(theta) * 10/cos(theta) - 4.9 * 100/cos^2(theta)
10 = 1000*sin(theta)*cos(theta) - 490/cos^2(theta) multiply through by cos^2 (theta)
10*cos^2(theta) = 1000*cos(theta)*sin(theta) - 490 That is really really ugly.

The only way I can see to solve this is to make a guess at the solution. Keep trying to narrow it down..

I'll try and get you an approximate answer.

I get 44.12 degrees roughtly or 0.7753 radians.

You can get t from t = 10/cos(theta)
t = 10 / cos(44.12) =13.92 seconds. You could refine the answer a bit by checking out the angle.

Report

New Topic

Home Search Q & A Board Gallery Resource Library Blog Dictionary Chat	Quick Links Start a Topic Latest Topics Unanswered Questions Courses Study Tools Browse by Textbook	Social Facebook YouTube Community Stats Who's Online Trending Staff	Help About Us Take the Tour Study Tips Posting Guidelines Terms and Policies Contact Us
Biology Forums - Study Force © 2010-2024 \| Powered by SMF \| SMF © 2015, Simple Machines \| Sitemap Biology-Forums.com is not affiliated with any publisher. Book covers, title and author names appear for reference only.