Answer to Question #79875 in Physics for Akhila

Question

A particle is falling freely under gravity . In first t seconds it covers distance x and in next t seconds, it covers distance y. Then t is given by
a)((y-x)/g)^1/2
b)((y+x)/g)^1/2
c)((2(y-x))/g)^1/2
e)((2(y+x))/g)^1/2

Answer 1

The distance covered in all time:
x+y=1/2 g(t+t)^2=1/2 g(2t)^2=1/2 4gt^2=2gt^2
t=(((x+y))/2g)^(1/2).
Answer: (((x+y))/2g)^1/2.