# Ramp Question Help.

Thread starter 7 years ago
#1
After a long evening of frustration, anger and self doubt i have decided to ask the collective intelligence of the student room in hope of helping me solve this annoying question.
" A stunt rider leaves the ramp at a 32 degree angle and lands at the same height after traveling a horizontal distance of 35m. Calculate his take off speed assuming negligible air resistance"
(the data in this question is altered as I'm looking for the method without cheating myself)
I would appreciate it very much if you guys can help me figure this out, its been a long couple of days and I' having a mental block.
0
7 years ago
#2
(Original post by tomeksoupforzepp)
After a long evening of frustration, anger and self doubt i have decided to ask the collective intelligence of the student room in hope of helping me solve this annoying question.
" A stunt rider leaves the ramp at a 32 degree angle and lands at the same height after traveling a horizontal distance of 35m. Calculate his take off speed assuming negligible air resistance"
(the data in this question is altered as I'm looking for the method without cheating myself)
I would appreciate it very much if you guys can help me figure this out, its been a long couple of days and I' having a mental block.
Using the constant acceleration questions for take-off speed v.

HORIZONTAL:

sx = 35
ux = vcos32
vx = vx
ax = 0 [negligible air resistance]
tx = t

VERTICAL:

sy = 0
uy = vsin32
vy = vy
ay = -g [acceleration due to gravity]
ty = t

sx = uxtx + ½axt2 ==> 35 = vtxcos32 ==> t = 35/(vcos32) (*)

sy = uy + ½ayt2 ==> 0 = vtsin32 - ½gt2 (**)

Substituting (*) into (**);

(35vsin32)/(vcos32)-½g(35/(vcos32))2 = 0

==> 35tan32 - ½g(1225/(v2cos232)) = 0

==> (1125g)/(2v2cos232)=35tan32

==> 2v2cos232=(1125g)/(35tan32)

==> v2 = (1125g)/(70tan32cos232)

==> v2 = (225g)/(14tan32cos232)

==> v =sqrt[((225g)/(14tan32cos232)]

Taking g = 9.80665 gives

v ≈ 18.7 ms-1 (3 s.f.)
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