Question

two particles are released from same height at an interval of 1 second. how long after the first particle begins to fall?? will the two particle be 10m apart? (g= 10m/s)

Answer 1

Task. Two particles are released from same height at an interval of 1 second. How long after the first particle begins to fall will the two particle be 10m apart? ($g=10~{}m/s^{2}$)

Solution. Notice that each particle moves zero initial velocity and constant acceleration $g=10~{}m/s^{2}$. Hence the distance passed by the first particle at time $t$ is given by the formula:

$h_{1}(t)=\frac{gt^{2}}{2}.$

The second particle is released at $t=1$ s, so the distance passed by the second particle at time $t\geq 1$ s is given by the formula:

$h_{2}(t)=\frac{g(t-1)^{2}}{2}.$

We should find $t$ such that

$h_{1}(t)-h_{2}(t)=10~{}m.$

Thus

$\frac{gt^{2}}{2}-\frac{g(t-1)^{2}}{2}=10$

$\frac{g}{2}\left(t^{2}-(t-1)^{2}\right)=10$

$t^{2}-t^{2}+2t-1=10\cdot\frac{2}{g}$

$2t=10\cdot\frac{2}{g}+1$

$t=\frac{10}{g}+\frac{1}{2}$

Substituting $g=10\,m/s^{2}$ we obtain that

$t=\frac{10}{10}+\frac{1}{2}=1+0.5=1.5~{}s.$