Question

A stone is dropped from a height 125m if g=10m- square what is the ratio of the distance travelled by it during the first and last second of its motion ​

Answer 1

Answer:

The question specifies g is 10 m/s^2.

In the first second, the stone travels S:

S = ut + 1/2 gt^2

S = 0t + 1/2 (10) 1^2

S = 5 meters.

Distance at the end of the first second is 5 meters.

The total number of seconds of fall is t:

S = ut + 1/2 gt^2

125 = 0t + 1/2 (10) t^2

25 = t^2

t = 5 seconds time of fall.

The distance traveled in the last second is the distance traveled by the end of the 5th second (125 meters) minus the distance traveled by the end of the 4th second.

4th second distance:

S = ut + 1/2 gt^2

S = 0t + 1/2 (10) 4^2

S = 5(16) = 80 meters

Distance at the end of 4 seconds is 80 meters.

Distance at the end of 5 seconds is 125 meters.

Last second distance = 125 meters - 80 meters = 45 meters

Ratio of first second distance to last second distance is 5 to 45.

Ratio is 1/9.

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