A body falling from rest was observed to fall through 78.4 m in 2 seconds. Find how long had it been falling before it was observed? (g = 9.8 m/s2)

To solve this problem, we first need to understand the physics behind it. The distance an object falls from rest is given by the equation d = 0.5gt^2, where g is the acceleration due to gravity and t is the time the object has been falling.

In this case, we know that the object fell 78.4 m in 2 seconds. So we can plug these values into the equation to find the value of g:

78.4 m = 0.5g(2 s)^2

Solving for g gives us g = 19.6 m/s^2. However, we know that the acceleration due to gravity is 9.8 m/s^2, not 19.6 m/s^2. This discrepancy tells us that the object must have been falling for some time before it was observed.

To find out how long the object had been falling before it was observed, we can set up the equation 0.5g(t + 2 s)^2 = 78.4 m + 0.5g(t)^2, where t is the time the object had been falling before it was observed.

Solving this equation for t gives us t = sqrt((78.4 m + 0.5g(t)^2) / 0.5g) - 2 s.

This equation is a bit tricky to solve, but with some algebraic manipulation, we can simplify it to t = sqrt((78.4 m / 0.5g) + t^2) - 2 s.

Finally, solving for t gives us t = sqrt((78.4 m / 4.9 m/s^2) + 4 s^2) - 2 s = sqrt(16 s^2 + 4 s^2) - 2 s = sqrt(20 s^2) - 2 s = 4 s - 2 s = 2 s.

So the object had been falling for 2 seconds before it was observed.