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- 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

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- anonymous

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- anonymous

The answer to the first part is 4.63.

- anonymous

just multiply 4.63 * 7037.6???

- 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

where 7037.6m/MIN is the conversion of 1520m/s

- anonymous

multiply that cos(theta) and flight time and you're good.

- anonymous

\[horizontal ~ range = v \cos(\theta) t_{flight}\]

- anonymous

Be sure to convert to km.

- anonymous

is theta 62.7 degrees.... i cant remember... :(

- anonymous

Yes.

- anonymous

ahh duh

- anonymous

still getting the wrong answer :/

- anonymous

wtf.

- anonymous

14.944 km.

- anonymous

7.04 km/min * cos 62.7 *4.63 min) = 14.944 km

- 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

ok I had v sub x and v sub y backwards.

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