question_answer

    A balloon going upward with a velocity of 12 m/s is at a height of 65 m from the earths surface at any instant. Exactly at this instant a ball drops from it. How much time will the ball take in reaching the surface of earth? \[(g=10m/{{s}^{2}})\]

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 \[h=65m,\] \[u=-12\text{ }m/s,\text{ }g=10\text{ }m/{{s}^{2}}\] \[\therefore \] \[65=-12t+\frac{1}{2}\times 10\times {{t}^{2}}\] \[\therefore \] \[5{{t}^{2}}-12t-65=0\] \[\Rightarrow \] \[(t-5)(5t+13)=0\] \[\therefore \]  \[t=5s\]