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Hot goIITian

Joined:	25 Jan 2009
Post:	176

24 Oct 2010 17:14:17 IST

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a particle is thrown with a velocity 40 m/s . if it passes A &B as shown in the figure at time t1 =1 sec &t2 = 3 sec . the value of h

ans) 15 m

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Comments (2)

Yagyadutt Mishra

Forum Expert

Joined: 19 Feb 2009

Posts: 1979

24 Oct 2010 17:29:44 IST

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It is very easy....

What you have given...that .

in time t= 1 sec ....ball is at height ....H

again in time t=3 ball is height ....H

Just write the two equation and solve....see..

The velocity responsible to move ball up to height H is the veritcal velocity...which usin(angle) = usin(30) = u/2

Hence vertical velocity = 40/2 = 20m/s

Now H = ut -1/2(g)(t)^2 { where g is always downward so taken as ...negative )

H = 20 (1) - 5(1) ^2 = 15 ( Put t=1)

H = 20(3) - 5(3)^2 = 15 (put t=3)

Hence the H = 15.....

~Áß???????

Blazing goIITian

Joined: 31 Aug 2008

Posts: 982

24 Oct 2010 20:44:45 IST

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WELL IN THAT QUES TOO MCH INFO IS GIVEN ,, U SHUDNT HV GIVEN U..THERE...WE CUD HV DONE THIS WITHOUT U ONLY!!!

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