The range of a projectile depends not only on v0 and θ0 but also on the value g of the free-fall acceleration, which varies from place to place. In 1936, Jesse Owens established a world's running broad jump record of 8.09 m at the Olympic Games at Berlin (g = 9.8128 m/s2)) Assuming the same values of v0 and θ0, by how much would his record have differed ( in cm) if he had competed instead in 1956 at Melbourne (g = 9.7999 m/s2)?
2006-09-28 16:12:48 · 5 answers · asked by mazoqo 1 in Science & Mathematics ➔ Physics
he moved in y axis for a certain time in order to get his maximum distance. so if the g is different the time travelled will be different therefor the maximum distance will change.
smaller value of g means greater distance, now:
Uy = g t
and Ux = DIstance / t time is the same though and we come to the point where
distance S1= Ux*Uy / g1 simiralrly
S2= Ux2 * Uy2 / g2
therefore
S2 distance in melbourne = S1*g1 / g2 =>
S2 = 8.1006m
that makes senses since it is greater for a smaller g.
therefore the difference is 1.065cm.
2006-09-28 19:05:28 · answer #1 · answered by Emmanuel P 3 · 1⤊ 0⤋
Hi. Good question but not that simple. His record was established at a certain temperature, altitude (oxygen content), humidity, etc. From a physics (rather than physical) standpoint, you should just set up a ratio. 8.10065 meters.
2006-09-28 16:16:07 · answer #2 · answered by Cirric 7 · 0⤊ 0⤋
Apparently you already know the formula, so you can determine the answer yourself by using it in both cases and comparing the difference. But of course, you then must presume all other factors are similar, which they were not.
2006-09-28 16:16:12 · answer #3 · answered by MrZ 6 · 0⤊ 1⤋
The first order aproximation is
8.09*(9.8128/9.7999 - 1)*100 = +1.06cm
2006-09-28 16:27:55 · answer #4 · answered by Steve 7 · 0⤊ 0⤋
u can caluculate this by (Vo^2SIN2(ANGLE)(diffs of g/product of g)
2006-09-28 17:17:44 · answer #5 · answered by venkatesh p 1 · 0⤊ 0⤋