- #1

Acnhduy

- 31

- 1

## Homework Statement

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.

## Homework Equations

Big 5

## 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 :)