Posted by intan on Sunday, March 14, 2010 at 11:03am.
(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)
Related Questions
Physics - A long jumper leaves the ground with an initial velocity of 12 m/s at ...
Physics - A jumper in the long-jump goes into the jump with a speed of 10m/s at ...
Physics - A jumper in the long-jump goes into the jump with a speed of 10m/s at ...
Physics..Urgent!!! - In the sport of pole vaulting, the jumper's point of ...
physics - A long jumper leaves the ground at an angle of 22.1◦ to the ...
physics - A boy throws a ball at 5.5 m/s at an angle of 25.4° above the ...
physics - A baseball is hit with a speed of 27 meters per second at an angle of...
PHYSICS - The object of the long jump is to launch oneself as a projectile and ...
PHYSICS - The object of the long jump is to launch oneself as a projectile and ...
PHYSICS - The object of the long jump is to launch oneself as a projectile and ...
For Further Reading
