## 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

1. anonymous

The answer to the first part is 4.63.

2. anonymous

just multiply 4.63 * 7037.6???

3. 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?

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

Be sure to convert to km.

8. anonymous

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

9. anonymous

Yes.

10. anonymous

ahh duh

11. anonymous

still getting the wrong answer :/

12. anonymous

wtf.

13. anonymous

14.944 km.

14. anonymous

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

15. 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]

16. anonymous

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