An archer lying on the ground shoots an arrow at a target that is 400 meters away, and 40 meters high. The initial velocity of the arrow is 120 m/s At what angle does he need to point his bow in order to hit the target? (Assume the value of g as -9.81 m/s² ).
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
Still confused. What am I solving for? Is Vi the initial velocity? And wouldn't it have two components: horizontal and vertical?
To find the angle at which the archer needs to point his bow, we can use the kinematic equations of projectile motion. The two main equations we will use are:
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.