Question

Find the ratio of the distance travelled by a freely falling body in the first second and third seconds of its fall

Answer 1

Answer:

1:5

Explanation:

I assume the 1st, 2nd & 3rd referred to here are seconds.

As we know, s = 0.5 *a*t^2, for a body with initial velocity=0

Though we know acceleration due to gravity g (which is our 'a') is known to be 9.8 m/sec^2, it would suffice to just use 'a' for acceleration, as we are intending to find out the ratio.

Distance traveled in 1 second = 0.5*a

Diatance traveled in 2 seconds = 0.5*a*2^2 = 2a

Distance traveled in 3 seconds = 0.5*a*3^2 = 4.5a

Now, distance traveled in 2nd sec = 2a - 0.5a = 1.5a

Distance traveled in 3rd sec = 4.5a-2a = 2.5a

The ratio of distances traveled 1st:2nd:3rd sec = 0.5:1.5:2.5 OR 1:3:5 (ANSWER)

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