Question

A person sitting in a helicopter, at 490 m height drops a stone. One sec later , he again throws a sec stone . both the stones hit the ground at the same time. Find out with what speed the person dropped the sec stone​

Answer 1

Answer:

in this question, there are two separate cases. but the common factor is the time . so, firstly we shall find out the time using the cases of first stone . once the time is abtained , we can put the time value for sec stone to calculate bin intinal velocity. Remember: first stone is dropped, so intinal velocity will be zero. sec stone bis thrown, so, intinal velocity will not be zero. Case1: dropping of first stone Given that ,. Intinal velocity of the first stone, u=0ms-1. Time taken= ts. Acceleration due to gravity, g =9.8ms-2. height from which the stone is dropped, s=490m from equ. of motion, = s= ut+1/2gtt. On putting values we have, 490=(0×t)+1/2(9.8×tt) tt= 590/4.9 =100 ,t =10. Cases