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HELP! A cannon tilted up at a 35 ∘ angle fires a cannon ball at 81 m/s from atop a 18-m-high fortress wall?
What is the ball's impact speed on the ground below? What is the ball's impact speed on the ground below? PLEASE HELP ASAP!!
- billrussell42Lv 73 months agoFavorite Answer
Vx = 81 cos 35 = 66.4 m/s
Vy = 81 sin 35 = 46.5 m/s
from the 18 m, how high does it go vertically
h = v²/2g = 46.5²/2•9.8 = 110.1 m
add 18 to that, total height = 128.1 m
speed after falling 128.1 m
v = √(2gh) = √(2•128.1•9.8) = 50.1 m/s
speed is vector sum horiz. and vert. speeds
v = √(66.4² + 50.1²) = 83.2 m/s
- oldschoolLv 73 months ago
Height h(t) = ho +Vyo*t - 4.9t² = 18 + 81sin35 *t - 4.9*t² = 0 at t = 9.854s
Vyf = Vyo - g*t = 81*sin35 - 9.8*9.854s = -50.11m/s
Vxf = Vxo = 81*cos35 = 66.35m/s
50.11² + 66.35² = 83.15² at arctan(-50.11/66.35) = -37° = 323°
Impact speed = 83m/s