Answer:
<h2>1.84feet</h2>
Step-by-step explanation:
Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;
Range ![R = U\sqrt{\frac{H}{g} }](https://tex.z-dn.net/?f=R%20%3D%20U%5Csqrt%7B%5Cfrac%7BH%7D%7Bg%7D%20%7D)
U is the velocity of the arrow
H is the maximum height reached = distance below the bullseye reached by the arrow.
R is the horizontal distance covered i.e the distance of the target from the archer.
g is the acceleration due to gravity.
Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?
On substitution,
![60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}](https://tex.z-dn.net/?f=60%20%3D%20250%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D%7D%20%5C%5C%5Cfrac%7B60%7D%7B250%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D%7D%5C%5C%5Cfrac%7B6%7D%7B25%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D)
Squaring both sides we have;
![(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet](https://tex.z-dn.net/?f=%28%5Cfrac%7B6%7D%7B25%7D%20%29%5E%7B2%7D%20%3D%20%28%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D%20%7D%20%29%5E%7B2%7D%20%5C%5C%5Cfrac%7B36%7D%7B625%7D%20%3D%20%20%5Cfrac%7BH%7D%7B32%7D%20%5C%5C625H%20%3D%2036%2A32%5C%5CH%20%3D%20%5Cfrac%7B36%2A32%7D%7B625%7D%20%5C%5CH%20%3D%201.84feet)
The arrow will hit the target 1.84feet below the bullseye.
Answer:
7,700
Step-by-step explanation:
350 x 22= 7,700
Answer:
Im pretty sure its 45 degrees
Step-by-step explanation:
Its an isosceles triangle so the upper 2 angles are Equal and a triangles angles adds up to 180 degrees, so if we have a 90 degree angle (the square) there's only 90 degrees left, so if you divide the 90 degrees by the remaining 2 Equal angles you get 45.