Answer:
tair = 4.81 s
vi = 45.6 m/s
θi = 76.8°
Step-by-step explanation:
In the x direction:
x₀ = 50 m
x = 0 m
v₀ = -vi cos θ
a = 0 m/s²
v = -vf cos 15° = -0.966 vf
Equation 1:
x = x₀ + v₀ t + ½ at²
0 = 50 + (-vi cos θ) t + ½ (0) t²
0 = 50 − vi cos θ t
Equation 2:
v² = v₀² + 2a(x − x₀)
(-0.966 vf)² = (-vi cos θ)² + 2 (0) (0 − 50)
0.966 vf = vi cos θ
vf = 1.035 vi cos θ
In the y direction:
y₀ = 0 m
y = 100 m
v₀ = vi sin θ
a = -9.8 m/s²
v = -vf sin 15° = -0.259 vf
Equation 3:
y = y₀ + v₀ t + ½ at²
100 = 0 + (vi sin θ) t + ½ (-9.8) t²
100 = vi sin θ t − 4.9t²
Equation 4:
v² = v₀² + 2a(y − y₀)
(-0.259 vf)² = (-vi sin θ)² + 2 (-9.8) (100 − 0)
0.067 vf² = vi² sin² θ − 1960
4 equations, 4 variables. Start by rearranging the first and third equations for vi cos θ and vi sin θ:
vi cos θ = 50 / t
vi sin θ = (100 + 4.9t²) / t
Substitute into the second and fourth equations:
vf = 1.035 (50 / t) = 51.76 / t
0.067 vf² = ((100 + 4.9t²) / t)² − 1960
Substitute the expression for vf and solve for t:
0.067 (51.76 / t)² = ((100 + 4.9t²) / t)² − 1960
179.49 / t² = (100 + 4.9t²)² / t² − 1960
179.49 = (100 + 4.9t²)² − 1960t²
179.49 = 10000 + 980t² + 24.01t⁴ − 1960t²
0 = 24.01t⁴ − 980t² + 9820.51
Solve with quadratic formula:
t² = 17.67 or 23.14
t = 4.20 or 4.81
If t = 4.20:
vi cos θ = 11.89
vi sin θ = 44.39
Divide second equation by first:
tan θ = 3.73
θ = 75.0°
vi = 46.0 m/s
Else, if t = 4.81:
vi cos θ = 10.39
vi sin θ = 44.36
Divide:
tan θ = 4.27
θ = 76.8°
vi = 45.6 m/s
Graph:
desmos.com/calculator/yvmr3tkp28
From the graph, we see the first solution is extraneous (the final θ is -15°).
So only the second solution is correct.