Could a person at the south pole see the north star, explain??
2 answers:
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Refer to the figure shown below, which is based on the given figure.
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
![4.9[ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ]^{2} - v_{0} sin(\theta - \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ] - X sin \alpha = 0](https://tex.z-dn.net/?f=4.9%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%20%5D%5E%7B2%7D%20-%20v_%7B0%7D%20sin%28%5Ctheta%20-%20%20%5Calpha%20%29%20%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%5D%20-%20X%20sin%20%5Calpha%20%20%3D%200)
Hence obtain
![aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta - \alpha )}]^{2} \\ b = cos \alpha \, tan(\theta - \alpha ) + sin \alpha](https://tex.z-dn.net/?f=aX%5E%7B2%7D-bX%3D0%20%5C%5C%20where%20%5C%5C%20a%3D4.9%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%5D%5E%7B2%7D%20%5C%5C%20%20b%20%3D%20cos%20%5Calpha%20%5C%2C%20%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%20%2B%20sin%20%5Calpha%20)
The non-triial solution for X is
Answer:
![X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}}](https://tex.z-dn.net/?f=X%3D%20%5Cfrac%7Bsin%20%5Calpha%20%20%2B%20cos%20%5Calpha%20%20%5C%2C%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%7B4.9%20%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20%5C%2C%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%20%20%5D%5E%7B2%7D%7D%20)
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer: X = 101.5 m
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
Hence obtain
The non-triial solution for X is
Answer:
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer: X = 101.5 m