Projectile motion question!

[Deceitful Wings]

How do I work this out?

[oppalovesme]

Not sure if this is right..

But the horizontal dispacement (x) = 10

and tan(20) = inital vertical velocity (u) / horizontal velocity (b)


b = 10 / total time taken (T)

let T = 2t

due to v (final vertical velocity, which is zero at midpoint) = u +at
0 = u - 9.8t
then u = 9.8t
hence t = u/9.8
therefore T = (2u)/9.8

since b = 10/T
b = 10/ ((2u) / 9.8 ) = 98/(2u)

therefore tan(20) = u / (98/(2u))
solving this, u ~ 4.22 m/s

using pythagoras, inital speed must be ~ 4.22/(sin(20))
which is ~ 12.35 m/s
hopefully this is right and makes sense D:

[Deceitful Wings]

yep, that's the answer! thanks for showing the working out I can't believe they would ask a question like this on a vcaa exam :S This was from vcaa 2004 lol

[oppalovesme]

Oh my gosh! I was actually tempted to write not to worry about it too much since it probably wouldn't appear on an exam
Hope there's nothing like that in this year's exam D:

[Phy124]

Not sure if this is right..

Looks good to me

An alternate method for anyone interested;





Vertical equation;



Displacement will be 0 (ending up at same vertical point - other side of jump), sub in



Horizontal equation;



Rearrange and you know that the horizontal distance is 10 and



Sub (2) into (1)



Simplify, rearrange etc.

[rayray17]

Don't know if this will help but this question will be so much easier done by using the range equation

(V^2(sin(2x20))/g=range where range equals 10 g=9.8

So you get sqrt(98/sin(40)) = v