A) 10 m/s
B) \[10\,m/s\]
C) \[\frac{40}{3}m/s\]
D) None of these
Correct Answer: C
Solution :
As seen from the cart, the projectile moves vertically upward and comes back. The time taken by cart to cover 80 m \[\frac{s}{v}=\frac{80}{30}=\frac{8}{3}s\] For a projectile going upward, \[a=-g=-10/m{{s}^{-2}},v'=0\] And \[t=\frac{8/3}{2}=\frac{4}{3}s\] \[\therefore \] \[v'=u+at\] Or \[0=u-10\times \frac{4}{3}\] Or \[u=\frac{40}{5}m{{s}^{-1}}\]You need to login to perform this action.
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