A long jumper's take-off velocity is 10 m/s at an angle of 20o to the horizontal. His centre
of gravity is 1 m above the ground at take-off and is 0.5 m above the ground at landing.
a)How far does he displace his centre of gravity horizontally between take-off and landing?
b) Is this the optimal angle of take-off? If not, what would the approximate optimal angle be
(no calculation necessary)? Given that this velocity vector is typical of world class long
jumpers, why do they not increase their angle of take-off? 
physic - drwls, Sunday, March 14, 2010 at 12:45pm
(a) Compute the time T that is takes to raise the C.G. from 1 m above ground to maximum height and back to 0.5 m. The time going up is t1 = Vo sin 20/g = 0.3486 s. The CG rises 0.5961 m in that time. To reach a point 0.5 m lower (at landing) requires time
t2 = sqrt(2*1.0961/9.81) = 0.4727 s
Thus T = t1 + t2 = 0.8213 s
The horizontal travel of the CG is
10*cos20*T = 7.72 m
(b) No. The optimum is closer to 45 degrees at that takeoff velocity.
(c) Taking off at a steeper angl, such as 40-45 degrees would require a large sudden loss of velocity because, at the moment of the jump, the angular momentum about the takeoff foot cannot change, and the velocity must drop by a larger amount than it would with a 20 degree takeoff. That would have a bad effect on the length of the jump.
Notice what happens, for example, to the veloity of a pole vaulter after he or she plants the pole. Their takeoff speed immediately becomes less than the previous runniung speed. (The highly flexible poles now used make the effect less noticeable)