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

smiley13

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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?

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tonyagodsey

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11 years ago

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n_m_m

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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

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darkshinra

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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.

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