A) 10 m/s
B) \[10\sqrt{8}\] 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 upwards and comes back. The time taken by cart to cover 80 m \[=\frac{s}{v}=\frac{80}{30}=\frac{8}{3}\,s\] Here, \[u=?,\,\,v=0,\,\,a=-g=10\,\,m/{{s}^{2}}\] (for a projectile going upwards) and \[t=\frac{8/3}{2}=\frac{4}{3}\,s\] From first equation of motion \[v=u+at\] \[0=u-10\times \frac{4}{3}\] \[u=\frac{40}{3}\,m/s\]You need to login to perform this action.
You will be redirected in
3 sec