UPSEE Solved Model Paper 7

As the stone falls from rest, its initial velocity is equal to zero, i.e. u = 0 Using the second equation of motion for a freely falling body (stone)r the distance covered by the stone in the first three seconds (t = 3 s) Is S = $ut+\frac{1}{2}at^2$ ⇒ $S_3$ = $0+\frac{1}{2}g(3s)^2$ ⇒ $S_3$ = $\frac{1}{2}\times 10 \text{m/s}^2 \times 9 \text{s}^2$ ⇒ $S_3$ = 45 m [For a freely falling body, a = g (acceleration due to gravity}. Also as the motion is only in one direction, i.e vertically downward, distance travelled is equal to displacement] Using $S_n$ = $u+\frac{1}{2}a(2n-1)$ , the distance covered in the last second (n = t) is $S_t$ = 0 + $\frac{g}{2}(2t-1)$ = $\frac{g}{2}(2t-1)$ ⇒ $S_t$ = $\frac{10 \text{m/s}^2}{2}$ (2t - 1) it is given that the distance covered by the stone in the last second is equal to the distance covered in the first three seconds of the motion. ⇒ $S_3$ = $S_t$ ⇒ 45 m = $\frac{10 \text{m/s}^2}{2}$ (2t - 1) ⇒ 2t - 1 = 9 ⇒ t = 5 s This is equal to the time for which the stone is in the air. Hence, option 'B' is correct