Question

A parachute going vertically upward with uniform acceleration 15.7 m/s². A stone is dropped from it. After 4 second another stone is dropped from it. Find the distance between the two stones 6 second after the the second stone is dropped.​

Answer 1

Explanation:

let for stone 1 t=0,u=0

for stone 2=4s

6s after

t=0+6=6s

t=6+4=10s

let position of s1 be ,x(6),s2be x(10)

distance betewen the stone is the difference beteween there position

s=ut+1/2at^2

s=x(10)-x(6)=100(15.7-4.9)-36(15.7-4.9)=691.2m

x(6)=0×6+1/2×9.8×36=g/2 × 36

net acceleration =15.7×36-4.9×36

x(10)=4.9×100-15.7×100

initial velocity for the 2 stones is 0