Practice question
Question
A stone is thrown upwards at 30m/s from a 55m tower. How far below the tower’s top is it after 7s? (Take g\=10m/s2)
Explanation
Displacement: y=30⋅7−12⋅10⋅(7)2=210−245=−35m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 35 m as the result, so option A is correct.