1no one tryin.......
someone pleaz answer these ques....
1) A bal is thrown vertically up wards. It is observed at a height h twice with a time interval t. The initial velocity of the ball is.......
2)A circus artist maintains 4 balls in motion . When one ball leaves his hand (speed = 20m/s) the position of other balls will be........
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8 Answers
62First one is not very difficult...
Let us say that the velocity at that height is v0.. then v0=g(t/2)
so we have velocity at height h
Now find the velocity at the initial time by using v02=u2+2as
where a = -g and s=h
[1]
622)A circus artist maintains 4 balls in motion . When one ball leaves his hand (speed = 20m/s) the position of other balls will be........
Find the total time for the balls to come down...
-20=20-gt
t=40/g = 4 seconds
So the interval of throwing the balls is 1 second..
Now solve the rest?
11) Use
h = ut - 1/2gt2
solving this for t, we have t=-2u±√(4u2-8gh)2g
these two values are the time interval betwn those points.....
so dt = t2-t1 = √(4u2-8gh)g =t
now get v....
1thank u all!!!!
but nishant sir pleaz explain how did u get V0=gt/2
62that is because the time to reach the top is t/2
and the velocity at the top is zero..
accelearation is g
so v=u+at
Put in the requisite values of v, u, a and t
1but sir how can we take time taken to reach the top =t/2 when it is given that t=\Deltat=t2-t1 where t1 & t2 are the two time intervals at which the ball is seen at height h???
also as u said putting v=0, t=t/2 , and a=-g in
v=u+at we get u=gt/2 i.e. the value of u=gt/2 but not V0=gt/2
PLEAZ EXPLAIN SIR..............