1018 one year ago The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 19 m/s. (a) What is the magnitude of the velocity of the projectile 1.7 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.7 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.7 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate

1. IrishBoy123

the first sentence tells you the projectile's horizontal velocity, which you will assume to be constant in this problem and that point you also know that its vertical velocity = 0 and that, more generally, acceleration in the vertical direction is -g. armed with that information, all you need are the equations of motion to solve https://gyazo.com/002724abd9cb7eb055122e07be79d344

2. 1018

thanks! im going to try it out

3. 1018

@Michele_Laino this one

4. 1018

i hate to ask, but can you also solve it? its worth alot of points on my homework and i hate to waste it

5. Michele_Laino

ok!

6. 1018

thank you so much!

7. Michele_Laino

here we have a parabolic motion, the situation is like below: |dw:1440003729324:dw|

8. 1018

oh wait! sorry!

9. 1018

The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the magnitude of the velocity of the projectile 1.2 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.2 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.2 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate 1.2 s after it reaches its maximum height?

10. Michele_Laino

at that point the kinetic energy is: $\Large KE = \frac{1}{2}m{v^2}$ where v is the velocity in my drawing, which is a horizontal vector

11. 1018

im sorry but it's this one. its actually the same but with different values sorry didnt see that

12. Michele_Laino

ok! I see

13. Michele_Laino

we have to compute the velocity when our projectile, passes at these points: |dw:1440004078499:dw|

14. Michele_Laino

namely at P1, and P2

15. 1018

ok. im following so far

16. Michele_Laino

here we have to keep in mind that the initial vertical velocity Vy is given by the subsequent formula: $\Large \frac{1}{2}mv_y^2 = mgh$ where h is the maximum height: |dw:1440004390763:dw|

17. Michele_Laino

|dw:1440004431440:dw|

18. 1018

what is m? and is g the gravitational acceleration?

19. Michele_Laino

m is the mass of our projectile, and g is gravity In a parabolic motion the horizontal velocity, doesn't change, so along all motion, the horizontal velocity is: $\Large {v_x} = 14m/\sec$

20. 1018

yes, and that's all i got so far. at least it's correct. haha

21. 1018

oh, i'll just see what you do. sorry im not familiar with using your formula, with the m one

22. Michele_Laino

the equation for the vertical component of the velocity is: $\Large {v_y}\left( t \right) = {v_{0y}} - gt$

23. Michele_Laino

where g is gravity and V0y is: $\Large {v_{0y}} = \sqrt {2gh}$ which comes from my formula above: $\Large \frac{1}{2}mv_y^2 = mgh$

24. Michele_Laino

so, after a substitution, we get: $\Large {v_y}\left( t \right) = {v_{0y}} - gt = \sqrt {2gh} - gt$

25. Michele_Laino

nevertheless we don't know the value of the maximum height h

26. 1018

yes

27. Michele_Laino

I think that we have insufficient data

28. Michele_Laino

we need to know, at least, the measure of the maximum height

29. 1018

sorry, but that is everything

30. Michele_Laino

I think that many parabolic motions are possible with the horizontal velocity which you have provided

31. 1018

someone answered this a while ago but i saw it late, maybe you can check out what he/she said

32. Michele_Laino

33. Michele_Laino

I think that with your data, the motion it is not uniquely determined

34. 1018

any other ideas how to do this?

35. Michele_Laino

I'm sorry, no my idea is that we need to know the maximum height or another parameter, for example the angle between the initial velocity and the horizontal line |dw:1440006104583:dw|

36. 1018

that's ok. thank you so much anyway. hey if you're interested, i could printscreen the solution. it shows the solution when after you answered incorrectly. haha

37. 1018

im gonna move on from this, try to get some rest, study again tomorrow. haha

38. Michele_Laino

39. 1018

ok, wait. i'll just make the answer wrong haha

40. Michele_Laino

ok!

41. 1018

here

42. 1018

it's a printscreen of the solution

43. Michele_Laino

yes! that is the solution, and I have failed, instead. Sorry!!

44. 1018

no no its ok. it's not like i figured it out too. haha. thank you so much anyway

45. Michele_Laino

ok! :)

46. 1018

hey, im gonna rest for awhile, i'll review this again tomorrow. haha. thank you again!

47. Michele_Laino

:)