the horizontal speed is 35 * cos30.0° = 30.3 m/s
So, it takes 47/30.3 = 1.55 seconds to travel.
So, plug that into the height
y = 1.3 + 35/2 t - 4.9t^2
to find the height at that time.
To find the correct distance, set
1.3 + 35/2 t - 4.9t^2 = 5.0
to find the time, then adjust the starting position.
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 35.0 m/s oriented 30.0° above the horizontal from a height of 1.30 m while standing 47.0 m away on your first try. You may neglect air resistance. What is the height of the arrow above the ground once it has travelled the 47.0 m horizontally?
If you fire at the same speed and angle on your second try, how far away could you stand to hit the orange? (Hint: There are actually two correct solutions for this. One is if you hit the orange while the arrow is still on its way up, the other is if you hit it while the arrow is coming back down. You need to find both.)
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