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anonymous
 4 years ago
An artillery shell is fired at an angle of 62.7
above the horizontal ground with an initial
speed of 1520 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell,
neglecting air resistance.
Answer in units of min
002 (part 2 of 2) 10.0 points
Find its horizontal range, neglecting air resis
tance.
Answer in units of km
anonymous
 4 years ago
An artillery shell is fired at an angle of 62.7 above the horizontal ground with an initial speed of 1520 m/s. The acceleration of gravity is 9.8 m/s2 . Find the total time of flight of the shell, neglecting air resistance. Answer in units of min 002 (part 2 of 2) 10.0 points Find its horizontal range, neglecting air resis tance. Answer in units of km

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer to the first part is 4.63.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just multiply 4.63 * 7037.6???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You need to find the horizontal velocity, which is\[v_x = v \cos(\theta)\]Then multiply this by the flight time. There is no way that the horizontal velocity can be greater than the total velocity. Where did you get 7037 from?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where 7037.6m/MIN is the conversion of 1520m/s

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0multiply that cos(theta) and flight time and you're good.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[horizontal ~ range = v \cos(\theta) t_{flight}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Be sure to convert to km.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is theta 62.7 degrees.... i cant remember... :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0still getting the wrong answer :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.07.04 km/min * cos 62.7 *4.63 min) = 14.944 km

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let's work through it. First, find the horizontal and vertical components of the velocity. \[v_x = 1520 \cos(62.7) ~ \left[\rm m \over s \right]\]\[v_y = 1520 \sin(62.7) ~ \left[\rm m \over s \right]\] The time of flight is\[t = {2 v_y \over g} ~ [\rm s]\] Distance traveled is\[d_x = v_x \cdot t ~[\rm m]\] Now, let's get some numbers. vx = 697.147 [m/s] vy = 1350.7 [m/s] t = 275.653 [s] dx = 697.147*275.653 = 192171 [m] We need the answer to be in km. Therefore, divide dx by 1000. dx = 192.171 [km]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok I had v sub x and v sub y backwards.
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