# Projectile Motion

**1. The problem statement, all variables and given/known data**

A man tosses object with initial velocity of 20 m/s into a room, and is 2.0 m above the ground. The room is 10.0 m above the ground.

In a provided diagram, the distance from man to room horizontally is 31.8 m.

I am required to solve for the angle of the object as it leaves the mans hand.

**2. Relevant equations**

Big 5

**3. The attempt at a solution**

v_{ix}= 20cosθ

v_{iy}= 20sinθ

To solve for θ, I must use the initial velocity given, broken up into components.

Since I am given the vertical displacement (8.0 m) and horizontal displacement (31.8 m) , I can use 2 formulas and substitute one into the other, by isolating t as it is common to both.

d_{x}= v_{ix}t

31.8 = 20cosθt

t= 31.8 / 20cosθ

d_{y}= v_{iy}t + 0.5a_{y}t^{2}

8.0 = 20sinθt – 4.9t^{2}

8.0 = 20sinθ(31.8 / 20cosθ) – 4.9(31.8 / 20cosθ)^{2}

8.0 = (31.8sinθ / cosθ) – 4.9(2.5281 / cos^{2}θ)

8.0 = (31.8sinθ / cosθ) – (12.39 / cos^{2}θ)

and now i’m stuck…. I tried moving to one side and changing to common denominator…

8.0 = (31.8sinθ / cosθ) – (12.39 / cos^{2}θ)

0 = (31.8sinθ / cosθ) – (12.39 / cos^{2}θ) – 8.0

0 = (31.8sinθcosθ – 12.39 – 8.0cos^{2}θ) / cos^{2}θ

0 = 31.8sinθcosθ – 12.39 – 8.0cos^{2}θ

0 = 15.9sin2θ – 12.39 – 8.0(1 – sin^{2}θ)

0 = 15.9sin2θ – 12.39 – 8.0 + 8sin^{2}θ

0 = 8sin^{2}θ + 15.9sin2θ – 20.39

but no luck…

i thought it looked like a quadratic at first, but I don’t think it works with sin2θ

any suggestions? thank you 🙂

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