Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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]

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.