question_answer

A ball is projected upwards from the top of tower with a velocity \[50\,\,m{{s}^{-1}}\] making an angle \[{{30}^{o}}\] with the horizontal. The height of tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground?

A)  2s           

B)  5s

C)  7s          

D)  9s

Correct Answer: C

Solution :

 If is time taken to reach the ground, then from equation of motion \[s=ut+\frac{1}{2}\,g{{t}^{2}}\].   Here, \[u=-v\sin {{30}^{o}}=-50\times \frac{1}{2}=-25\,\,m/s\] \[g=10\,m/{{s}^{2}},s=70\,\,m\] \[\therefore \] \[70=-25\,t+\frac{1}{2}\times 10\times {{t}^{2}}\] \[\Rightarrow \] \[5\,{{t}^{2}}-25\,t-70=0\] \[\Rightarrow \] \[t=-2\,\,s\] or \[t=7\,\,s\] Since, - 2 s is not valid, therefore \[t=7\,s\]