Question

10. A stone falls freely such that the distance
covered by it in the last second of its motion
is equal to the distance covered by it in the first
5 seconds. It remained in air for :
1) 12 s 2) 13 s 3) 25 s 4) 26 S
​

Answer 1

Answer:25 seconds

Explanation:

Answer 2

Answer:

Stone remained in the air for 13 s

Explanation:

Initial velocity of the stone u =0 m/s

Distance covered in first 5 seconds is

$S=ut+\frac{1}{2} at^2\\S=0\times 5+\frac{1}{2} \times 9.8 \times 5^2\\S=122.5 m$

Let time for which stone remained in the air is t sec

Distance covered by stone in the last second of motion is

$S=S_{t}-S_{t-1}\\S=ut+\frac{1}{2} at^2-[u(t-1)+\frac{1}{2} a(t-1)^2]\\S=\frac{1}{2} at^2-\frac{1}{2} a(t-1)^2]\\S=\frac{1}{2} at^2-\frac{1}{2} at^2-\frac{a}{2} +at\\S=-\frac{a}{2} +at\\122.5=\frac{-9.8}{2} +9.8 \times t\\t=13 s$

Stone remained in the air for 13 s