A = angle up
u = 120 cos A
400 = u t = 120 cos A * t
t = 400 / (120 cos A)
Vi = 120 sin A
v = Vi - 9.81 t
40 = Vi t -4.9 t^2
u = 120 cos A
400 = u t = 120 cos A * t
t = 400 / (120 cos A)
Vi = 120 sin A
v = Vi - 9.81 t
40 = Vi t -4.9 t^2
1. Vertical motion equation: Δy = v₀y * t + (1/2) * g * t²
2. Horizontal motion equation: Δx = v₀x * t
Where:
- Δy is the vertical displacement (40 meters in this case).
- Δx is the horizontal displacement (400 meters in this case).
- v₀y is the vertical component of the initial velocity (which we need to determine).
- v₀x is the horizontal component of the initial velocity (which is known, 120 m/s).
- g is the acceleration due to gravity (-9.81 m/s²).
First, let's find the time it takes for the arrow to reach the target horizontally. We can use the horizontal motion equation:
Δx = v₀x * t
Rearranging the equation:
t = Δx / v₀x
t = 400 m / 120 m/s
t ≈ 3.33 s
Now, let's find the vertical component of the initial velocity, v₀y, using the vertical motion equation:
Δy = v₀y * t + (1/2) * g * t²
Rearranging the equation:
v₀y = (Δy - (1/2) * g * t²) / t
v₀y = (40 m - (1/2) * (-9.81 m/s²) * (3.33 s)²) / 3.33 s
v₀y = (40 m - (1/2) * (-9.81 m/s²) * 11.0889 s²) / 3.33 s
v₀y = (40 m + (1/2) * 551.97 m²/s²) / 3.33 s
v₀y = (40 m + 275.985 m²/s²) / 3.33 s
v₀y ≈ 125.081 m/s
Now that we have the vertical component of the initial velocity (v₀y) and the horizontal component (v₀x = 120 m/s), we can find the angle (θ) at which the archer needs to point his bow using the following trigonometric equation:
tan(θ) = v₀y / v₀x
Substituting the values:
tan(θ) = 125.081 m/s / 120 m/s
tan(θ) ≈ 1.042
Now, we can solve for θ by taking the inverse tangent (arctan) of both sides:
θ = arctan(1.042)
θ ≈ 47.65°
Therefore, the archer needs to point his bow at an angle of approximately 47.65° in order to hit the target.