A) 5s
B) 5 s
C) 10s
D) None of these
Correct Answer: A
Solution :
From equation of motion, we have \[h=ut+\frac{1}{2}g{{t}^{2}}\] where u is initial velocity, g the acceleration due to gravity and t the time taken . Taking upward direction as negative and downward direction as positive, we have \[h=65m,\] \[u=-12\text{ }m/s,\text{ }g=10\text{ }m/{{s}^{2}}\] \[\therefore \] \[65=-12t+\frac{1}{2}\times 10\times {{t}^{2}}\] \[\therefore \] \[5{{t}^{2}}-12t-65=0\] \[\Rightarrow \] \[(t-5)(5t+13)=0\] \[\therefore \] \[t=5s\]You need to login to perform this action.
You will be redirected in
3 sec