A body falls from rest. In the last second of its fall it covers half of the total distance. If g is 9.8 text m/s2, then the total time of its falls is (in second)

Question

A body falls from rest. In the last second of its fall it covers half of the total distance. If g is 9.8 text m/s2, then the total time of its falls is (in second) (A) 2 (B)  2+√2  (C)  4-√2  (D) 3

Tardigrade Admin · Accepted Answer

According to equation h=(1/2)gt2 (h/2)=(1/2)g(t-1)2 Solving Eqs. (i) and (ii), we get t=2± √2s t=2+√2s

Q. A body falls from rest. In the last second of its fall it covers half of the total distance. If g is $9.8 m / s^{2},$ then the total time of its falls is (in second)

2

$2 + 2$

$4 - 2$

3

According to equation $h = \frac{1}{2} g t^{2}$
$\frac{h}{2} = \frac{1}{2} g (t - 1)^{2}$
Solving Eqs. (i) and (ii), we get
$t = 2 \pm 2 s$
$t = 2 + 2 s$

Q. A body falls from rest. In the last second of its fall it covers half of the total distance. If g is 9.8m/s2, then the total time of its falls is (in second)

2

2+2​

4−2​

3

According to equation h=21​gt2 2h​=21​g(t−1)2 Solving Eqs. (i) and (ii), we get t=2±2​s t=2+2​s

Q. A body falls from rest. In the last second of its fall it covers half of the total distance. If g is $9.8 m / s^{2},$ then the total time of its falls is (in second)

$2 + 2$

$4 - 2$

According to equation $h = \frac{1}{2} g t^{2}$
$\frac{h}{2} = \frac{1}{2} g (t - 1)^{2}$
Solving Eqs. (i) and (ii), we get
$t = 2 \pm 2 s$
$t = 2 + 2 s$