anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
The answer to the first part is 4.63.
anonymous
  • anonymous
just multiply 4.63 * 7037.6???
anonymous
  • anonymous
You 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?

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anonymous
  • anonymous
where 7037.6m/MIN is the conversion of 1520m/s
anonymous
  • anonymous
multiply that cos(theta) and flight time and you're good.
anonymous
  • anonymous
\[horizontal ~ range = v \cos(\theta) t_{flight}\]
anonymous
  • anonymous
Be sure to convert to km.
anonymous
  • anonymous
is theta 62.7 degrees.... i cant remember... :(
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
ahh duh
anonymous
  • anonymous
still getting the wrong answer :/
anonymous
  • anonymous
wtf.
anonymous
  • anonymous
14.944 km.
anonymous
  • anonymous
7.04 km/min * cos 62.7 *4.63 min) = 14.944 km
anonymous
  • anonymous
Let'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
  • anonymous
ok I had v sub x and v sub y backwards.

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