At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

|dw:1343399513614:dw|

Show that \(R = 2H\)

well how do i start? anyone able to give me a hint?

i dont know vectors..

|dw:1343400018031:dw|

well what do i do next? I am unable to find the trajectory equation thing for \(B\)

what about the velocity?

i am able to find the acceleration, velocity for \(A\) but not \(B\)

no ... i'm asking about the initial velocity of both projectiles

i dont know; 0?

what's this R anyway?

R is the range; the longest horizontal distance travelled by the object

|dw:1343402352605:dw|
abit clearer; sorry for the horrible drawing

trivial case v=0, R=0, H=0 ... what' the initial velocity of both projectiles .. are they same?

well the velocity i got for A for the x-motion is \(u\) and the y-motion is \(-gt\)

what about the B?

i dont know how to do B

because its projected vertically so im lost

i mean what is the initial velocity of B ... going up?

i dont know..i guess its going up..

how did you get y = (1/2)gt^2 for the second equation?

v^2 = y^2 + 2as => H = u^2/2g

okay; thanks!

thank you!

yw