Jacaranda question i need help with plz (1 Viewer)

[elfanger]

Pg 33 question 9b - Jacaranda new edition book

A cyclist wants to perform a stunt in which he rides up a ramp,
launching himself into the air, then flies through a hoop and lands on another
ramp. the angle of each ramp is 30 degrees and the cyclist is able to reach the
launch height of 1.50m with a launching speed of 30km/h

Calculate:

How far away
the landing ramps should be placed?

I tried to use delta x = ut and i found the distance of x to be 3.02 isnt that the answer? or am i not getting somthing here?

 

[Xayma]

What is the answer? I got 6.855m from the end of the ramp to the end of the other ramp. (ie the straight down drops are facing into the middle)

 

[:: ck ::]

vy= uy + at
0 = 4.165 + (-9.8)t (uy derived from usin@)
t = 0.425s
trip time = 2t = 2 x 0.425 = 0.85s
ux = 8.3cos30 = 7.19ms-1

delta u = uxt = 7.19 x 0.85 = 6.13 m

 

[:: ck ::]

by the way, i got the answers from taking my workbook out

dunno if the questoin is the same... it sounds like it is so yeh...

i think the reason y ur answer is small is coz u didn multiply t times 2

 

[elfanger]

Thanks alot guys. The answer is 6.13 m
Your right, when i measured t, i measured it with v=0 hence only measuring it to its maximum height, and i forgot to multiply by 2.

Thanks again.

 

[:: ck ::]

no probz

multiplying t by 2 is very easy to forget coz the t only accounts for half the vertical trip going up (from bottom to max height)

itz very easy to just go straight into delta u = uxt using the wrong value

 

[Xayma]

Bah I knew I rounded too early (then when I played around with it I rounded that off again etc). Hmm I should of done it that way, easier,

 

[victorling]

jacaranda just got answers for its numerical questions...i think physics in contexts is better in content although does not have any solution to its questions ...