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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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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.....