A brick falling freely from a helicopter drops 40 meters during a certain 1-second time interval. The distance it will fall in the next second is A. 56 B. 45 C. 80 D. 50
but if we take account of the distance it has travelled in 1 sec it would define its velocity..i.e if its travelling 40 m in 1 sec then its initial velocity could be taken as u=40m/s \[a=10m/s ^{2}\] t=1 sec then acc to relation \[s=ut+1/2at ^{2}\] we would get \[s=40(1)+1/2\times10(1)^{2}=40+5=45m\]
that's why i am considering the initial velocity u to be 40m/s as v is changing wrt to acceleration(gravity) at 10m/s^2..and measuring the distance regarding it!!
i have considered v as final velocity that depend on time duration at which we are supposed to calculate it..
i have not considered the average velocity i have only considered the initial velocity that we require in the relation \[s=ut+1/2at ^{2}\] where u=initial velocity
so are we here looking for the distance in nth second or n seconds..??
I finally get what you're saying. Yes, the answer is 50m.
i got it the answer is 50...!!..thanks @demitris
Join our real-time social learning platform and learn together with your friends!