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Stone on Roof, Find Distance

  • Thread starter postfan
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  • #1
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Homework Statement


We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 44 ∘. The gravitational acceleration is g=10 m/s2. (See figure)

(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)

(b) What time does it take the stone to reach the roof? (in seconds)

Homework Equations





The Attempt at a Solution


I set up 2 equations :
s+15=20*cos(44)*t
s/sqrt(3)+6=20*sin(44)*t-5t^2
and got s=.99 and t=1.11. What did I do wrong?
 

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  • #2
Simon Bridge
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Homework Statement


We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 44 ∘. The gravitational acceleration is g=10 m/s2. (See figure)

(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)

(b) What time does it take the stone to reach the roof? (in seconds)

Homework Equations





The Attempt at a Solution


I set up 2 equations :
s+15=20*cos(44)*t
s/sqrt(3)+6=20*sin(44)*t-5t^2
and got s=.99 and t=1.11. What did I do wrong?
How did you arrive at those equations?
(Please show your reasoning.)
How did you account for the slope of the roof?
 
  • #3
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I used the equation x=v_0*t+.5*a*t^2. I created a 30-60-90 triangle , and realized that for each distance s horizontally it goes up by s/sqrt(3). The total distance horizontally is 15+s and vertically is (s/sqrt(3)_6) and we are given the acceleration and launch velocity/angle.
 
  • #4
Simon Bridge
Science Advisor
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As a description, that leaves a lot to be desired.

You need to know where the trajectory parabola of the stone intersects the line of the roof.
What is your strategy for figuring that out?
 

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