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Emergency!


A 2 meter tall archer shoots an arrow with the speed of 20 m/s at a target positioned on the same horizon plane 30 m away. If the arrow is to hit the center of the vertical target, which is 2 meters above the ground at what angle to the horizontal should the arrow be launched. [Given sin2a = 2sina.cosa]

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egenriether
If I understand the question the arrow starts at 2m above ground and will try to hit a target 2m above ground 30m away. So if this is the case we can neglect the 2m altogether. It starts at some height (which we can call 0) and after its flight it lands at the same height. So we need to solve its vertical and horizontal position functions to find the angle. The vertical function will tell us how long it's in the air before it gets back to the height it started at. We then have to ensure the angle allows it to travel 30m horizontally in that time.
Vertical: a=-9.81m/s^2
v=-9.81t + 20sin(n). Where n is the angle
So vertical position is the integral of velocity
s=-9.81/2 t^2 +20t sin(n)
s=-4.905t^2+20t sin(n)=0
This can help us find n in terms of t. We need t such that the arrow traveling at 20m/s makes it exactly 30m. Using the formula s=vt
30=20t cos(n)
t=1.5/cos(n)
Putting this in our previous equation
-4.905(1.5/cos(n))^2+
20(1.5/cos(n))sin(n)=0
Now multiply both sides by (cos(n))^2 to get
-4.905(1.5)+20(1.5)cos(n)sin(n)=0
Or
-7.36+30sin(2n)=0
Now find n
7.36= 30sin(2n)
7.36/30=sin(2n)
2n=arcsin(7.36/30)
2n=14.2
n=7.1 degrees

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