Question

William Tell is said to have shot an apple off his son’s head with an arrow. If the arrow was shot with an initial speed of 50.0 m/s and the boy was 28.0 m away, at what launch angle above the horizontal did Bill aim the arrow? (Assume that the arrow and apple are initially at the same height above the ground. Hint: sin2θ = 2sinθcosθ)

Answer 1

here

Vx = 50cosθ. This is because Vx/50 = cosθ, so we can solve for the horizontal component.
Vy = 50sinθ. This is because Vy/50 = sinθ, so we can solve for the vertical component.

From here, we can see that t * 50cosθ = 28 meters or that t= 28/50cosθ 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. 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=28/50cosθ)

We can now plug in for a, Vo and t to solve for θ.


0= (1/2) (-9.8 m/sec^2) (28/50cosθ)^2 + 50sinθ (28/50cosθ)

0= -4.9(28/50cosθ)^2 + 28sinθ/cosθ

0= -1.53(1/(cosθ)^2) + 28tanθ

0= -1.53(secθ)^2 +28tanθ

0= -1.53(1 - (tanθ)^2) + 28tanθ

0= -1.53 - 1.53 (tanθ)^2 + 28tanθ

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 = 0.0548

θ= 3.137 degrees