4. An archer shoots an arrow at 75.0 m distant target; the bull's-eye of the target is at the 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? b. There is a large tree halfway between the archer and the target. It is 3.50 m above the release height of the arrow. Will the arrow go over or under the tree? 4Equation Sheet One Dimensional Motion: Projectile Motion Trigonometry Ax = X - XO x = Xo + Vxt tand = At = ty - to y = yo +- (Voy + vy) cost = 2 Av 0= 310 At Vy = Voy - gt sind = Ax 1= At y = yo + Voyt -79tz x = Xo + 0t vy = Voy - 2g(y - yo) Vo + v D= g = +9.8m/s2 2 v = Vo + at R - Vosin200 x = Xo + Vat + = at2 v2 = v3 + 2a(x - X0) Hypotenuse (H Opposite (0) Angle Adjacent (A]]Answer all questions. Show how you arrived at your answers. Remember to include units where appropriate. Pay attention to significant figures