# Projectile Motion man tosses object

1. Feb 26, 2014

### Acnhduy

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

vix= 20cosθ
viy= 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.

dx= vixt
31.8 = 20cosθt
t= 31.8 / 20cosθ

dy= viyt + 0.5ayt2
8.0 = 20sinθt - 4.9t2
8.0 = 20sinθ(31.8 / 20cosθ) - 4.9(31.8 / 20cosθ)2
8.0 = (31.8sinθ / cosθ) - 4.9(2.5281 / cos2θ)
8.0 = (31.8sinθ / cosθ) - (12.39 / cos2θ)

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

8.0 = (31.8sinθ / cosθ) - (12.39 / cos2θ)

0 = (31.8sinθ / cosθ) - (12.39 / cos2θ) - 8.0

0 = (31.8sinθcosθ - 12.39 - 8.0cos2θ) / cos2θ

0 = 31.8sinθcosθ - 12.39 - 8.0cos2θ

0 = 15.9sin2θ - 12.39 - 8.0(1 - sin2θ)

0 = 15.9sin2θ - 12.39 - 8.0 + 8sin2θ

0 = 8sin2θ + 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 :)

2. Feb 26, 2014

### Simon Bridge

... so you have interpreted the question as asking about the direction of the initial velocity vector, given the ball passes through points A and B (going from A to B) and you know the speed at A.

... trig identities not helping you - you may be able to do it geometrically seeing $2\sin^2\theta = 1-\cos2\theta$ ... or is there another kinematic equation you haven't considered yet - that may be called into play?
i.e. have you used all the information at your disposal yet?

Notes:
do not put the numbers in until the end or you'll get turned around.
keep track of your reasoning all the way through.

3. Feb 26, 2014

### Simon Bridge

Hmmm ... perhaps:
... go back to where it looks like you almost have a quadratic in $\cos\theta$,
... multiply through by $\sec^2\theta$ ... from there, Pythagoras should get you a quadratic in $\tan\theta$.

4. Feb 26, 2014

### Acnhduy

Mhm. Didnt even notice that it could be changed to secant. Thankyou

5. Feb 27, 2014

### Simon Bridge

No worries.
The tan vs sec identities crop up a lot - especially in exams.