A hill has an angle of elevation of 13.5°. From a point on the
hill, a ball is thrown upward with an initial velocity of 80
ft/sec and an angle of elevation of 45°. Neglecting air
resistance, find the distance down the hill to the point of
impact.
Just Show me how to start it; you don't have to do the whole thing. I know it is something that has to do with x=Vocosθt and y=-16t^2 + Vosinθt + ho
hill, a ball is thrown upward with an initial velocity of 80
ft/sec and an angle of elevation of 45°. Neglecting air
resistance, find the distance down the hill to the point of
impact.
Just Show me how to start it; you don't have to do the whole thing. I know it is something that has to do with x=Vocosθt and y=-16t^2 + Vosinθt + ho
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You are indeed correct about x and y.
Let the point on the hill where the ball was thrown be origin. Then h₀ = 0
v₀ = initial velocity = 80 ft/sec
θ = angle ball was thrown = 45°
x(t) = v₀ cosθ t = 40√2 t
y(t) = −16t² + v₀ sinθ t + h₀ = −16t² + 40√2 t
We express y in terms of x:
t = x/(40√2)
y = −x²/200 + x
You don't say if person throwing the ball is facing up or down the hill, but since you ask to find the distance down the hill to the point of impact, I assume this person is facing down the hill. So hill slopes down (i.e. negative slope)
Equation of the hill: y = mx
where m = slope of the hill = ∆y/∆x = −tan(13.5°)
y = −tan(13.5°) x
Now you just have to find x- and y-coordinates of point where parabola (trajectory of ball) intersects line (equation of slope). Once you've found this point, use distance formula from origin to this point to find distance down the hill to point of impact.
y = −tan(13.5°) x
y = −x²/200 + x
−tan(13.5°) x = −x²/200 + x
Here's a diagram that illustrates trajectory of ball and slope of the hill.The two red points are the point where ball was thrown and the point where ball lands on the hill.
http://www.wolframalpha.com/input/?i=%E2…
Exact coordinates of point of impact:
(200(1+tan(13.5°)), −200tan(13.5°)(1+tan(13.5°)) = (248.02, −59.54)
Distance from origin to point (248.02, −59.54) = 255.06
Let the point on the hill where the ball was thrown be origin. Then h₀ = 0
v₀ = initial velocity = 80 ft/sec
θ = angle ball was thrown = 45°
x(t) = v₀ cosθ t = 40√2 t
y(t) = −16t² + v₀ sinθ t + h₀ = −16t² + 40√2 t
We express y in terms of x:
t = x/(40√2)
y = −x²/200 + x
You don't say if person throwing the ball is facing up or down the hill, but since you ask to find the distance down the hill to the point of impact, I assume this person is facing down the hill. So hill slopes down (i.e. negative slope)
Equation of the hill: y = mx
where m = slope of the hill = ∆y/∆x = −tan(13.5°)
y = −tan(13.5°) x
Now you just have to find x- and y-coordinates of point where parabola (trajectory of ball) intersects line (equation of slope). Once you've found this point, use distance formula from origin to this point to find distance down the hill to point of impact.
y = −tan(13.5°) x
y = −x²/200 + x
−tan(13.5°) x = −x²/200 + x
Here's a diagram that illustrates trajectory of ball and slope of the hill.The two red points are the point where ball was thrown and the point where ball lands on the hill.
http://www.wolframalpha.com/input/?i=%E2…
Exact coordinates of point of impact:
(200(1+tan(13.5°)), −200tan(13.5°)(1+tan(13.5°)) = (248.02, −59.54)
Distance from origin to point (248.02, −59.54) = 255.06