Question

a freely falling covers 0.1m in 0.1sec and 0.2m in next 0.1sec. find the value of acceleration due to gravity at the place.

Answer 1

Task. A freely falling covers 0.1 m in 0.1 sec and 0.2 m in next 0.1 sec. Find the value of acceleration due to gravity at the place.

Solution. Assume that the acceleration $g$ due to gravity is constant at the place, so the object moves with constant acceleration. Let $v$ be its initial velocity. Then the distance passed by time $t$ is equal to

$h(t)=vt+\frac{gt^{2}}{2}.$

By assumption,

$h(0.1\ sec)=0.1\ m.$

Moreover, in next 0.1 sec., that is at $t=0.1+0.1$ the distace was

$h(0.1+0.1)=h(0.2)=0.1+0.2=0.3$

So we get the following system of equations:

$h(0.1)=0.1=v*0.1+\frac{g*0.1^{2}}{2},$ $h(0.2)=0.3=v*0.2+\frac{g*0.2^{2}}{2}$

$0.1v+0.005g=0.1,$ $0.2v+0.02g=0.3.$

$0.2v+0.01g=0.2,$ $0.2v+0.02g=0.3.$

Subtracting left equation from the right one we get

$0.2v+0.02g-0.2v-0.01g=0.3-0.2$

$0.01g=0.1$

$g=\frac{0.1}{0.01}=10\ m/s^{2}.$

Answer. $g=10\ m/s^{2}$.