Dynamics, Angle of projectile (1 Viewer)

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This problems has given me a lot of difficulty even asking several people. Kinematics of Particles using 2D vectors.

v(initial) = 400 m/s. Determine two angles of elevation (theta) which will permit the projectile to hit the mountain target at B = (5 km)i+(1.5 km)j.

wont directly let me host url because i'm new sorry. pm if picture needed thanks all.
 

tiny-tim

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Welcome to PF!

Hi whtan20! Welcome to PF! :smile:

With difficult questions, go one step at a time …

First step: what equations do you know that might help here? :smile:
 
I know that
vx0=400cos(theta)
vy0=400sin(theta)

and integrating..v=dx/dt

x = 400tcos(theta)
y = 400tsin(theta)

and from here i'm clueless as where to go. Thanks.
 
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x = v0*cos(θ)*t, so t = x/(v0*cos(θ))
y = v0*sin(θ)*t - 1/2 g*t^2,
y = x*tan(θ) - 1/2 g*x^2/(v0*cos(θ))^2
use 1/(cos(θ))^2 = (sec(θ))^2 = 1 + (tan(θ))^2
y = x*tan(θ) - 1/2 g*x^2/v0^2 - 1/2 g*x^2*(tan(θ))^2/v0^2
then we get a quadratic equation for tan(θ) that is, (after plug all known values)
765.63(tan(θ))^2 - 5000tan(θ) + 2265.63 = 0
use quadratic formula and get
tan(θ) = 6.0407, so θ = 80.6 degrees
tan(θ) = 0.48987, so θ = 26.1 degrees
 
Thanks a lot Simon, i did basically exactly that, but didn't know how to turn it into a quadratic, too many terms all over the place.
 

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