please help me in solving the problem
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I'm not sure bout the ans
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Hi friend !!
In the given question,u=0 since initial velocity of ball before throwing it up is 0
a= g=10m/s^2
Time =t=10secs
So velocity with which it was thrown up = v= u+at
=0+10X10
=100m/s
Maximum height attained =s=ut+1/2at^2
=100 x 10+1/2 x (-10) x 10 x 10
(Please note that the negative sign of g is taken here because the ball is moving up against the acceleration.
So a= g= -10m/s^2)
=1000-500
=500m
So the height attained is 500m.
Hope that clears your doubt friend !
In the given question,u=0 since initial velocity of ball before throwing it up is 0
a= g=10m/s^2
Time =t=10secs
So velocity with which it was thrown up = v= u+at
=0+10X10
=100m/s
Maximum height attained =s=ut+1/2at^2
=100 x 10+1/2 x (-10) x 10 x 10
(Please note that the negative sign of g is taken here because the ball is moving up against the acceleration.
So a= g= -10m/s^2)
=1000-500
=500m
So the height attained is 500m.
Hope that clears your doubt friend !
QUANTUM900:
I think anomi is ryt .
u will be 50 m/sec
if we will take the initial point as A and the highest point as B then the ball motion is from A to A . And by putting the values in s = uT + aT^2 ÷ 2. we will get 'T' which is total time =2u÷g and from here we will get u=50 m/sec and s= 125 m
OK koi baat nhi late hi sahi ;)
Hmm......
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