Physics Question??

[Krakerjak]

An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper?

Answer is 11 m/s.

This question has me stumped, i can't get the right answer.

I start off by getting the max vertical distance

TAN(23) x 4.35 = 1.85m

find the time in air y= -1/2at^2

square root of -(2 x 1.85)/-9.81 m/s^2) = 0.614s

velocity in x direction is constant so velocity is 4.35m/.614s = 7.08 m/s

velocity at 23 degrees = 7.08 m/s / cos(23) = 7.7 m/s?? should be 11 m/s
can anyone see my error, i don't imagine the text is wrong.

 

[ApacheXMD]

"TAN(23) x 4.35 = 1.85m"

don't think that's right

edit: the vertical component of the initial velocity is Vsin23. So you use that to find time in air (kinematics equation, keep it in terms of V)
Knowing time of flight, and distance, you can find the magnitude of V

-patchy

 

[Dolemite69]

I know I don't know... But here, ask Britney!!!

Britney's Guide to Semiconductor Physics

 

[GoodToGo]

If u take 8.7 instead of 4.35, u get the rite answer, but I cant back it with any logic

 

[ThisIsMatt]

got time?

 

[Krakerjak]

You're right goodtogo, hmm... but i can't see why you would take the full vertical distance if you are trying to find the max hight which is at the midpoint....oh well. As long as i don't forget that on my midterm tomorrow

 

[ApacheXMD]

by taking (tan23)(.5)(8.7) to be your max verical distance, you're assuming the dude is jumping like this /\
like two straight lines, one going upwards, and one coming back down.

you can't just take tan23(4.35) for the max height. it's wrong.

-patchy

 

[GoodToGo]

edit: nm, stupid post

 

[stingrae]

For starters u are wrong abt the max vertical height coz ur treating the problem like as if the jumper takes a triangular path..in fact he takes a parabolic path...now as this holds true...i have a couple of equations for this so called projectile ..
the horizontal distance travelled = [u^2sin(2x23.5)]/g ...where g is acceleration due to gravity and u is initial takeoff speed...
u know horizontal distance travelled = 8.7m and g=9.8m/s^2
doing this u'll get the right answer...

if u want i'll derive the whole equation.....but it'll take a lot of time

 

[SupaDupaCheez]

Why don't you ask MrFrog840....he'll give his retarded friend a buck for the answer

Link

Just a little humor in a very heavy thread (physics never was my strong point)

SDC

 

[SupaDupaCheez]

Bump for a fixed link

SDC

 

[Krakerjak]

thx, i got it. Just couldn't grasp the theory of how the problem was solved.