particle is dropped from height of 20 m from ground there is wind blowing due to which horizontal acceleration of the particle becomes 6 metre per second square .find the horizontal displacement of the particle till it reaches the ground
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Displacement=12m
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sourabh28:
you got 50 points for free
In S=ut+1/2at^2
ut will be zero
ut will be zero
Thanks
i ve got the answer from internet
yes
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Your questions is nice just wait two minutes The answer will be
S= 20
U=10
S=ut+1/2at*2
20=1/2*10*t*2
T*2=20/5
T*2=4
T=2
It is displaced for 2 sec
S= ut +1/2 at*2
S= 1/2 *6*4
S=12 M
The horizontal distance will be 12 m
S= 20
U=10
S=ut+1/2at*2
20=1/2*10*t*2
T*2=20/5
T*2=4
T=2
It is displaced for 2 sec
S= ut +1/2 at*2
S= 1/2 *6*4
S=12 M
The horizontal distance will be 12 m
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