Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A 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]

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.