- #1

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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.

- Thread starter whtan20
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- #1

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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.

- #2

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Last edited by a moderator:

- #3

tiny-tim

Science Advisor

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Hi whtan20! Welcome to PF!

With difficult questions, go one step at a time …

First step: what equations do you know that might help here?

- #4

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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.

- #5

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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

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