Answer:
After 0.19288 sec second object hit the ground after hitting first object
Explanation:
We have given two objects are dropped from at the same time from height 0.49 m and 1.27 m
As the object is dropped so its initial velocity will be zero u = 0 m/sec
For object 1 height h = 0.49 m
From second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]0.49=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.31622 sec
For second object
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]1.27=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.5091 sec
So difference in time = 0.5091-0.31622 = 0.19288 sec
So after 0.19288 sec second object hit the ground after hitting first object
