A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

jiteshmeghwal9

  • one year ago

Path of a projectile projected from the ground is given as \[y=ax-bx^2\]. Find angle of projection, maximum height attained and horizontal range of the projectile.

  • This Question is Closed
  1. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Michele_Laino

  2. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @mukushla

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You can sole this with formulas your teacher provided, but I prefer solve it mathematically

  4. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The problem is that the question has given me only one equation & has asked me to find three things.

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well note that the trajectory of projectile is a parabola, can you plot that parabola for me?

  6. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1435902087173:dw|

  7. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where I have written [\a\]

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you have \[y=ax-bx^2\]sry I mean \(b\)

  9. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[b\] must be a +ve number because it is a parabolic equation or we can say quadratic

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well \(b>0\) because parabola must be downward, right? It's a ball moving on the ground \(x-axis)

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can this be a path for projectile?|dw:1435902758838:dw|

  12. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How can the body move against gravity?

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, so \(b\) can not be less than zero, because if \(b<0\) we will have a upward parabola

  14. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now, parabola (ball) hits the x-axis (ground) at \(y=0\)\[y=ax-bx^2=0\]\[x=0 \]\[x=\frac{a}{b}\]

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1435903099764:dw|

  17. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1435903307350:dw|

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    xm is where maximum height (ym) occurs

  19. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now if you think about it mathematically and let \(f(x)=ax-bx^2\): for angle of projection: \(\tan \theta= f'(0)\) maximum height attained: \(y_m\) and horizontal range of the projectile: \(x=\frac{a}{b}\)

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[y_m=f(x_m)\]

  22. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is \[f'(0)\] ?

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    derivative

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    have you studied derivative?

  25. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nope not yet

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, so how you find it? do you have formula or something?

  27. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have eqn of trajectory\[y=xtan \theta [1-\frac{ x }{ R }]\] 'R' is range.

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks, equate \(f(x)\) and formula and find \(R\) and \(\theta\)

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[y=\tan \theta \ x - \frac{\tan \theta}{R} x^2=ax-bx^2 \]\[\tan \theta =a\]\[\frac{\tan \theta}{R}=b\]

  30. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so\[\frac{ \tan \theta }{ \frac{ \tan \theta }{ R }}=\frac{ a }{ b }\]\[R=\frac{ a }{ b }\]also\[\tan \theta=a\]\[\theta =\tan^{-1} a\]

  31. jiteshmeghwal9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    got it, thanks friend :)

  32. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.