question_answer

A stone is projected up from the top of a tower 58.8 m high with a velocity of 19.6 m/s. In what time will it reaches the foot of the tower?

Answer:

If h is the height of the tower, displacement\[=-h\] So using \[\therefore \,s=ut+\frac{1}{2}a{{t}^{2}}\] \[-h=19.6\,\,t-\frac{1}{2}\times 9.8\,{{t}^{2}}\] Since\[u=19.6\,m{{s}^{-1}}\]and it moves up initially. i.e. \[-58.8=19.6\,\,t-4.9\,\,{{t}^{2}}\] \[4.9\,{{t}^{2}}-19.6\,\,t-4.9\,\,{{t}^{2}}-58.8={{t}^{2}}-4t-12=0\] \[\Rightarrow (t-6)(t+2)=0\,\,\,\therefore \,\,t=6s\]