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bazaltina [42]
3 years ago
7

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?
Physics
1 answer:
mart [117]3 years ago
4 0

Answer:

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

Explanation:

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

<em>For horizontal component, we will have:</em>

<em>ν cos(θ)t = x ⇒ cos(θ) =  x/νt</em>

<em>For vertical component we will have:</em>

<em>h= v sin(θ)t ₋ gt²÷ 2 ⇒ sin (θ) = h + gt²÷2/νt</em>

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

<em>tan(θ) = h +gt²÷2/νt/ x÷νt = h+ gt²÷2/x = 3+ 9.8.0.5²÷2/20 = 0.21125</em>

<em>Now we can calculate θ = tan⁻¹(0.21125) ≈ 11.9°</em>

<em>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:</em>

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

<em></em>

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