Question

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Answer 1

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

Answer 2

Answer:

a) Θ = 18.5°

b) h = 6.26 m, 3.50 m, arrow goes over the branch

Explanation:

having the following data:

Vo = 35 m/s

Θ = ?

horizontal distance = 75 m

a)

using the following equations:

Voy = Vo*(sin Θ) = 35*(sin Θ)

Vox = Vo*(cos Θ) = 35*(cos Θ)

horizontal distance to target = 75 = Vox*(2t); where t = Voy/g

replacing values:

75 = Vox*(2/g)*(Voy) = 2*(Vox)*(Voy)/9.8 =(Vo)²*[2*(sin Θ)*(cos Θ)]/g = (Vo)²(sin2Θ)/g

solving and using trigonometric identities:

sin2Θ = 75*(g)/(Vo)² = 75*(9.8)/(35)² = 0.6

2Θ = 36.91°

Θ = 18.5°

b)

The time to reach the maximum height will be equal to:

t = Voy/g = 35*(sin18.5°)/9.8 = 1.13 s

and the maximum height will be equal to:

h = 1/2gt² = (0.5)*(9.8)*(1.13)² = 6.26 m, 3.50 m, arrow goes over the branch