Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

the equation of projectile is y = ax - bx2 . its horizontal range is?

Sagot :

Everest2017
y = a x - b x^2
Range is a/b

y = tan Ф x -  g x²  / 2 u² cos² Ф
tan Ф = a        -  equation 1
b = g / 2u² cos² Ф          so  u² cos² Ф = g /2b    - equation 2

R = u cos Ф * 2 * u sin Ф / g  = 2/g  sinФ  u² cos Ф
     = 2 /g   tan Ф   u² cos² Ф            by using equation 1  and equation 2
     = (2 /g )  a  (g / 2b ) = a / b

Answer: Range, [tex]R =\frac{a}{b}[/tex]

Explanation:

The equation of trajectory is:

[tex]y = x tan \theta (1-\frac{x}{R})[/tex]

Where, [tex]\theta[/tex] is the angle of projectile, R is the horizontal range.

The equation of projectile is:

y = ax-bx²

[tex]\Rightarrow y = ax(1-\frac{b}{a}x)[/tex]

On comparing:

[tex]tan \theta = a[/tex]

[tex]R = \frac{a}{b}[/tex]

Hence, the horizontal range is [tex]R =\frac{a}{b}[/tex]

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.