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.

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_ixt

31.8 = 20cosθt

t= 31.8 / 20cosθ

d_y= v_iyt + 0.5a_yt²

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

You have missed out one piece of information - the relationship between x and y on impact.

 

This appears as my third equation.

 

You appear to like messing around with trig, so have a go yourself and post agian if you need further help.

 

Edited February 27, 201411 yr by studiot

If I see such exercises, I always use what seems the most general formula for a projectile's trajectory to me:

[math]y=y_0+x\cdot\tan{\theta}-\frac{g\cdot x^2}{2v^2\cos^2{\theta}}[/math]

 

Where [math]y[/math] is the height of landing, [math]y_0[/math] is the initial height, [math]\theta[/math] is the angle at which the projectile is thrown, [math]x[/math] the horizontal displacement of the projectile, [math]g[/math] the earth's gravity constant [math](\approx 9.8 ms^{-2}[/math] and [math]v[/math] the initial velocity of the projectile.

 

So what I would do: use what studiot told you and then use this most wonderful formula.

Edited February 28, 201411 yr by Function