question_answer

A ball is projected upwards from the top of a tower with a velocity of 50 m/s making in angle of \[30{}^\circ \] with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing, will the ball reach the ground? (Take\[g=10\,m/{{s}^{2}}\])

A)   2s                  

B)  5 s

C)   7s                  

D)  9 s

Correct Answer: C

Solution :

Vertical component of initial velocity \[=50\sin {{30}^{o}}\]\[=25\,m/s.\] \[U\sin \,g\,s=ut+\frac{1}{2}a{{t}^{2}}\]along the y-axis \[-70=25t-\frac{1}{2}\times 10\times {{t}^{2}}\] or \[5{{t}^{2}}-25t-70=0\] or \[{{t}^{2}}-5t-14=0\] or \[t(t-7)+2(t-7)=0\] or \[t=7\,\sec .\] Hence, the correction option is [c].