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
[tex]0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta - \alpha ) \frac{d}{u} - h = 0 [/tex]
[tex]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[/tex]
Hence obtain
[tex]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 [/tex]
The non-triial solution for X is
[tex]X= \frac{b}{a} [/tex]
Answer:
[tex]X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}} [/tex]
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
