The distance s in metres covered by a body, dropped from a tall cliff, as a function of time t in seconds is given by

Question

The distance s in metres covered by a body, dropped from a tall cliff, as a function of time t in seconds is given by (A) s=9.8 t2 (B) s=9.8 t (C) s=4.9 t2 (D) s=4.9 t

Tardigrade Admin · Accepted Answer

We know that, s=u t+(1/2) g t2, where s = distance, u= initial velocity, t= time, g - gravitational acceleration Here, u=0, g=9.8 m / s 2 s=0 × t+(1/2) × 9.8 t2 s=49 t2 m text in t s

Q. The distance $s$ in metres covered by a body, dropped from a tall cliff, as a function of time $t$ in seconds is given by

$s = 9.8 t^{2}$

$s = 9.8 t$

$s = 4.9 t^{2}$

$s = 4.9 t$

We know that,
$s = u t + \frac{1}{2} g t^{2}$ , where
$s$ = distance, $u =$ initial velocity, $t =$ time,
$g$ - gravitational acceleration
Here, $u = 0, g = 9.8 m / s^{2}$
$s = 0 \times t + \frac{1}{2} \times 9.8 t^{2}$
$s = 49 t^{2} m in t s$

Q. The distance s in metres covered by a body, dropped from a tall cliff, as a function of time t in seconds is given by

s=9.8t2

s=9.8t

s=4.9t2

s=4.9t