A cannon fires a shell with a speed of 84m/s .when the cannon is inclined at 45° ,the horizontal distance covered is observed as630m .what is the percentage decrease in the horizontal distance observed due to air resistance
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Vo = 84m/s[45o].
Dx = Vo^2*sin(2A)/g=84^2*sin90/9.8=720 m
with zero wind.
(720-630)/720 * 100% = 12.5% Decrease.
Vo = 84m/s[45o].
Dx = Vo^2*sin(2A)/g=84^2*sin90/9.8=720 m
with zero wind.
(720-630)/720 * 100% = 12.5% Decrease.
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Answer:
decrease in percentage =12.5%
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