need clear explanation You would like to shoot an orange out of a tree with your

Question

need clear explanation

You would like to shoot an orange out of a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. You fire the arrow at 32.0 m/s oriented 30.0degree above the horizontal from a height of 1.20 m while standing 58.0 m away on your first try. You may neglect air resistance. What is the height of the arrow once it has travelled the 58.0 m horizontally? If you fire at the same speed and angle on your second try, how far away can you stand to hit the orange?

Explanation / Answer

vx = V*cos(30) = 27.7 m/s
vy = v*sin(30) = 16 m/s

t = x/vx = 58/27.7 = 2.094 s


h = 1.2 + vy*t - 0.5*g*t^2 = 13.2 m


5 = 1.2 + vy*t - 0.5*g*t^2


4.9*t^2 - 16*t + 3.8 = 0


solving above eqn we get

t = 0.2578 s


x = Vx*t = 27.7 * 0.3 = 7.14 m

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