<span>Given:
Hmax (distance) = 50.0m
v</span>₀ = <span>70.0m/s
Required:
what angle should the arrow make with the horizontal as it is being shot
Solution:
Hmax = v</span>₀²sin²θ / 2g
sin²θ = 2gHmax / v₀²
sin²θ = 2 (9.81 m/s²) (50m) / (70 m/s)²
sin²θ = 0.200
θ = 26.56°
The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.
Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.
r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6
v - 232.5 m/s
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