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?

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

the cellphone, they would then know if the killer would have contacted them or always hated them. or if it was a breakup. but if that is not the case, then it would be the clothing. to put in a bullet, it would have to be put in by hands. hands have finger prints which, would have giving you the answer of who done it.

explanation:

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge