## idku one year ago Math-Based-Physics question.

1. yolomcswagginsggg

9 + 10?

2. anonymous

19

3. yolomcswagginsggg

21.

4. idku

A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s. -------------------------------------------------------------

5. idku

STOP IT, GOD DAMN IT !

6. yolomcswagginsggg

xD

7. idku

PART 1: 1) What is the initial speed of the cannonball? $$\sqrt{33^2(m/s)+25^2(m/s)~}=41(m/s)$$

8. idku

PART 2: 2) What is the initial angle θ of the cannonball with respect to the ground?PART 1: $$\theta=\tan^{-1}\left(25/33\right)\approx37.15$$

9. idku

PART 3 3) What is the maximum height the cannonball goes above the ground?

10. idku

$$\Large v^2=u^2+2ax$$ (NOTE: x is vertical displacement) Ok, so I know that v=0 when the ball reaches maximum height. My initial vertical velocity is 33m/s. Acceleration is -9.81a $$\Large 0^2=33^2+2(-9.81)x$$ $$\Large 33^2=2(9.81)x$$ $$\Large x=55.5$$ and that is wrong...

11. idku

I just want to do it a plain way using some familiar formula

12. idku

oh, it should be 25 not 33

13. idku

because we are computing vertical displacement we must use vertical velocity

14. idku

$$\Large v^2=u^2+2ax$$ (NOTE: x is vertical displacement) Ok, so I know that v=0 when the ball reaches maximum height. My initial vertical velocity is 25m/s. Acceleration is -9.81a $$\Large 0^2=25^2+2(-9.81)x$$ $$\Large 25^2=2(9.81)x$$ $$\Large x=31.8$$ and that is CORRECT!

15. idku

Okay, I am getting through it with a painful struggle-:(:)

16. ganeshie8

17. idku

Like nothing else in this world.... :) But, true that

18. idku

Well, the fact that my teacher is awful gives me more credit

19. idku

4) How far from where it was shot will the cannonball land?

20. idku

$$\Large x = ut + \dfrac{1}{2}at^2$$ $$\Large 31.8 = 25t - \dfrac{9.81}{2}t^2$$ t = 2.44229 t = 2.65455 then I can use: Horizontal displacement = time × initial horizontal velocity my problem now is which time to take

21. ganeshie8

you should not get multiple "t" values the vertex is a "single" point |dw:1441337672713:dw|

22. idku

31.855249745=25t−(9.81/2)t^2

23. idku

then I get about same thing 2.54 seconds http://www.wolframalpha.com/input/?i=31.855249745%3D25t%E2%88%92%289.81%2F2%29t%5E2

24. ganeshie8

use the exact value for $$x$$, then u will get a single $$t\ value 25. ganeshie8 use the exact value for \(x$$, then u will get a single $$t$$ value

26. idku

well, I got a single approximation for t (as for now because I increased my value-precision)

27. idku

2.54 seconds

28. ganeshie8

looks fine

29. idku

(Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = (2.54s) × (33m/s)=83.82s but the practice thingy tells me that is wrong

30. idku

oh wait 2.54 is half of the whole time, right?

31. idku

Yup. (Horizontal displacement) = (time) × (initial horizontal velocity) (Horizontal displacement) = $$\color{red}{\rm 2 \times }$$ (2.54s) × (33m/s)=167.6m

32. idku

BINGO

33. ganeshie8

Yes, 2.54 = time at which the particle reaches max height

34. idku

Yup. Like that $$\Downarrow$$ really (Horizontal displacement) = ($$\color{red}{\rm 2 \times }$$ 2.54s) × (33m/s)=167.6m

35. idku

Yes, so from max back to v=0 another 2.54s

36. idku

Thank you!

37. idku

And lets see If I can also bear through question 5. ------------------------------------------ 5) What is the speed of the cannonball 3.7 seconds after it was shot?

38. idku

I know that $$\rm t = 3.7(s)-2.54(s)=1.16(s)$$ $$\rm u = 41(m/s)$$ --------------------- I am trying to use $$\rm v=u+at$$ but I keep arriving on wrong results... lets see...

39. idku

oh, I also tried t=3.7 without subtracting and it was still wrong