Saturday, November 14, 2009

While sitting on a tree branch 10.0 m above the ground, you drop a chestnut.?

While sitting on a tree branch 10.0 m above the ground, you drop a chestnut. When the chestnut has fallen 2.2 m, you throw a second chestnut straight down. What initial speed must you give the second chestnut if they are both to reach the ground at the same time?

While sitting on a tree branch 10.0 m above the ground, you drop a chestnut.?
time taken for 1st chestnut to fall through remaining 7.8 m = time taken by 2nd chestnut to fall 10m





time taken for 1st nut to fall 7.8 m :


= time taken for it to fall 10m - time taken for it to fall 2.2 m


= sqrt(2 S/g) - sqrt (2s/g)


= sqrt (2*10/10) - sqrt(2*2.2/10)


= sqrt(2) - sqrt(0.44)


= 1.41 - 0.66


= 0.75 seconds





Now this time must be the time taken for 2nd nut to fall 10 m:


so....


s = ut + 1/2*g.t^2


10 = u(0.75) + 1/2*10*(0.75)^2


10 = 0.75 u + 2.8125


0.75 u = 7.1875


u = 7.1875 / 0.75


u = 9.58 m/s approx.
Reply:What Are You Talking About?
Reply:AFTER 2.2 m the droped one has to travel 10-2.2=7.8m .the time it takes to reach the ground=time taken by the thrown chestnut..as the droped chest travels 2.2 m under free fall ,hence it gains some velocity after faliing 2.2 m..the veocity after fallin 2.2 m can be calculated by using the eqation V(square)-u (square)=2*a*s,,where u=0,a=9.8,s=2.2 after getting the velocity at that point again use the formula v=u+at, where u is the calculated value above.get the time taken from this eqation..that maenas this much time should be taken by the thrown kajju.then use the formula, s=ut+0.5*a*t(square).put s=10,a=9.8,t=calculated avobe then find u..which is the required answer of yr question.


No comments:

Post a Comment