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.

see earlier