William Tell is said to have shot an apple off of his son's head with an arrow. If the arrow was shot with an initial speed of 55 m/sec and the boy was 15 meters away, at what launch angle did William aim the arrow?
We can start by drawing the vector for the velocity of the arrow. Because the initial velocity is 55 m/sec, the magnitude of the vector is 55. The angle is unknown. The resulting vector (and its horizontal and vertical components) are pictured below.
Vx = 55cosθ. This is because Vx/55 = cosθ, so we can solve for the horizontal component.
Vy = 55sinθ. This is because Vy/55 = sinθ, so we can solve for the vertical component.
We can now draw position time, velocity time, and acceleration time graphs for the motion in both the horizontal and vertical directions.
Horizontally, the motion is not affected by any opposing forces, so the arrow will exhibit constant velocity. This is represented by an upward sloping line because the arrow is moving forward at a constant speed. This also translates to a constant function in the velocity time graph and no acceleration because acceleration is the slope of the velocity time graph and the graph has a slope of 0. The graphs for the horizontal direction are pictured below.
Vy = 55sinθ. This is because Vy/55 = sinθ, so we can solve for the vertical component.
We can now draw position time, velocity time, and acceleration time graphs for the motion in both the horizontal and vertical directions.
Horizontally, the motion is not affected by any opposing forces, so the arrow will exhibit constant velocity. This is represented by an upward sloping line because the arrow is moving forward at a constant speed. This also translates to a constant function in the velocity time graph and no acceleration because acceleration is the slope of the velocity time graph and the graph has a slope of 0. The graphs for the horizontal direction are pictured below.
For the vertical direction, we must consider the effects of gravity, namely, acceleration due to gravity. Due to this acceleration, the object will initially travel up, then return down, giving it a parabolic trajectory. Due to this, the velocity time graph will be linear with a negative slope, showing how the arrow will slow as it reaches its peak and then speed up in the downward direction. The initial velocity is 55sinθ (vertical component) and the object has a constant acceleration of -9.8 m/s^2. The resulting graphs are pictured below.
We can now begin to solve the problem. The other piece of information that is given is that the boy is 15 meters away. This means that the horizontal displacement Δx is equal to 15 meters. We can show this in the horizontal velocity time graph, because displacement is the area under the velocity curve. We can construct a relation using this fact that can be used later.
From here, we can see that t * 55cosθ = 15 meters or that t= 15/55cosθ seconds.
At this same time t, the vertical displacement should be zero (Due to the fact that the apple and the arrow launch point are at the same height). We can use the general equation for an object with constant acceleration, which is what is exhibited by the vertical motion.
At this same time t, the vertical displacement should be zero (Due to the fact that the apple and the arrow launch point are at the same height). We can use the general equation for an object with constant acceleration, which is what is exhibited by the vertical motion.
y = (1/2)at^2 + Vot + yo
And since y- yo is equal to Δy, we can say that:
0= (1/2)at^2 + Vot (at time t=15/55cosθ)
We can now plug in for a, Vo and t to solve for θ.
0= (1/2) (-9.8 m/sec^2) (15/55cosθ)^2 + 55sinθ (15/55cosθ)
0= -4.9(15/55cosθ)^2 + 15sinθ/cosθ
0=-4.9 (225/3025(cosθ)^2) + 15tanθ
0= -0.3645(1/(cosθ)^2) + 15tanθ
0= -0.3645(secθ)^2 +15tanθ
0= -0.3645(1 - (tanθ)^2) + 15tanθ
0= -0.3645 - 0.3645 (tanθ)^2 + 15tanθ
Thus, the solutions to the quadratic equation above will give us θ. There are two values, but one will be negative and can be discarded.
Solving for tanθ:
tanθ = 0.0243
θ= 1.39 degrees
And since y- yo is equal to Δy, we can say that:
0= (1/2)at^2 + Vot (at time t=15/55cosθ)
We can now plug in for a, Vo and t to solve for θ.
0= (1/2) (-9.8 m/sec^2) (15/55cosθ)^2 + 55sinθ (15/55cosθ)
0= -4.9(15/55cosθ)^2 + 15sinθ/cosθ
0=-4.9 (225/3025(cosθ)^2) + 15tanθ
0= -0.3645(1/(cosθ)^2) + 15tanθ
0= -0.3645(secθ)^2 +15tanθ
0= -0.3645(1 - (tanθ)^2) + 15tanθ
0= -0.3645 - 0.3645 (tanθ)^2 + 15tanθ
Thus, the solutions to the quadratic equation above will give us θ. There are two values, but one will be negative and can be discarded.
Solving for tanθ:
tanθ = 0.0243
θ= 1.39 degrees
William Tell would have to shoot the arrow at 1.39 degrees above the horizontal to knock the arrow off of his son's head.