An arrow has an initial launch speed of 32 m/s. If it must strike a target 38 m away at the same elevation, what should be the projection angle?
Range = Vo^2*sin(2A)/g = 38
32^2*sin(2A)/9.8 = 38
104.5*sin(2A) = 38
sin(2A) = 0.36364
2A = 21.3o
A = 10.7o
To calculate the projection angle of the arrow, we can use the equations of projectile motion.
First, let's break down the initial velocity of the arrow into its horizontal and vertical components. Since the arrow is launched with an initial speed of 32 m/s, we can determine its vertical component and horizontal component using trigonometry.
The vertical component of the initial velocity (Vy) can be calculated using the formula:
Vy = V * sin(θ)
where V is the initial speed of the arrow and θ is the projection angle.
Similarly, the horizontal component of the initial velocity (Vx) can be calculated using the formula:
Vx = V * cos(θ)
where V is the initial speed of the arrow and θ is the projection angle.
Since the arrow must strike a target 38 m away, we can use the equation of motion for the horizontal direction:
Δx = Vx * t
where Δx is the horizontal displacement and t is the time of flight.
In this case, the horizontal displacement (Δx) is given as 38 m.
Since there is no vertical displacement (Δy = 0), we can determine the time of flight (t) using the vertical equation of motion:
Δy = Vy * t + (0.5) * g * t^2
where Δy is the vertical displacement and g is the acceleration due to gravity (-9.8 m/s^2).
In this case, the vertical displacement (Δy) is 0, so the equation simplifies to:
0 = Vy * t + (0.5) * g * t^2
Now, let's solve these equations to find the projection angle (θ).
First, we solve the vertical equation for t:
0 = Vy * t + (0.5) * g * t^2
0 = (V * sin(θ)) * t - (4.9) * t^2
Rearranging the equation:
(4.9) * t^2 = (V * sin(θ)) * t
Divide both sides by t:
4.9 * t = V * sin(θ)
Now, we solve the horizontal equation for t:
38 = Vx * t
Substitute the expressions for Vx and t:
38 = (V * cos(θ)) * (4.9 * t / (V * sin(θ)))
Now, substitute the value of t from the second equation into the first equation:
4.9 * t = V * sin(θ)
4.9 * (38 / (V * cos(θ))) = V * sin(θ)
Simplify the equation:
4.9 * 38 = V^2 * sin(θ) * cos(θ)
Substitute the value of V^2 * sin(θ) * cos(θ) from the first equation:
4.9 * 38 = 4.9 * t * cos(θ)
Cancel out the common terms:
38 = 4.9 * cos(θ)
Now, solve for cos(θ):
cos(θ) = 38 / (4.9)
θ = arccos(38 / 4.9)
Using a calculator, we find:
θ ≈ 73.24 degrees.
Therefore, the projection angle should be approximately 73.24 degrees.