This time, William Tell is shooting at an apple that hangs on a tree ( Fig. 3.32). The apple is a horizontal distance of 20.0 m away and at a height of 4.00 m above the ground. If the arrow is released from a height of 1.00 m above the ground and hits the apple 0.500 s later, what is the arrow's initial velocity?

The initial arrow's velocity is 40,9 m/s at 11.9° from the horizontal

Explanation:

In order to find the inital velocity we need to determine its components and the angle that the arrow is launched at.

For horizontal component, we will have:

ν cos (θ) t = x ⇒ cos (θ) = x/νt

For vertical component we will have:

h = v sin (θ) t ₋ gt²: 2 ⇒ sin (θ) = h + gt²:2/νt

From the two equations we got, after noting that the vertical displacement is 3m, we can calculate

tan (θ) = h + gt²:2/νt / x:νt = h + gt²:2/x = 3 + 9.8.0.5²:2/20 = 0.21125

Now we can calculate θ = tan⁻¹ (0.21125) ≈ 11.9°

Now that we know the angle we can subtitute at any of the expressions for the two components of the velocity. Let's do this subsitution at the horizontal component:

ν cos (θ) t = x = ν = x/tcos (θ) = 20 / 0.5cos (11.9) ≈ 40.9 m/s.