Question

. A boy on a 20 m high cliff drops a stone. One second later, he throws down another stone such that
both the stones hit the ground simultaneously. Find the initial velocity of the second stone. (g = 10 ms-2
)

Answer 1

Task. A boy on a $h=20$ $m$ high cliff drops a stone. One second later, he throws down another stone such that both the stones hit the ground simultaneously. Find the initial velocity of the second stone. ($g=10$ $m/s^{2}$)

Solution. The first stone moved with initial zero velocity and acceleration $g=10$ $m/s^{2}$. Therefore its height at time $t$ is given by the formula

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

The time $T$ when he reach the ground satisfies the following equation:

$h_{1}(T)=0=h-\frac{gT^{2}}{2},$

whence

$\frac{gT^{2}}{2}=h$

$T=\sqrt{\frac{2h}{g}}.$

Substituting the values we get

$T=\sqrt{\frac{2*20}{10}}=\sqrt{4}=2\ s.$

The second stone moved with some initial velocity $v$ and the same acceleration $g=10$ $m/s^{2}$. Since it starts one second later, its height at time $t$ is given by the formula

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

Since both stones reach the ground simultaneously, we have also that

$h_{2}(t)=0=h-v(T-1)-\frac{g(T-1)^{2}}{2},$

whence

$v(T-1)=h-\frac{g(T-1)^{2}}{2}$

$v=\frac{h-\frac{g(T-1)^{2}}{2}}{T-1}$

$v=\frac{h}{T-1}-\frac{g(T-1)}{2}$

Substituting the values of $h$, $g$, and $T$ we get

$v=\frac{h}{T-1}-\frac{g(T-1)}{2}=\frac{20}{2-1}-\frac{10*(2-1)}{2}=20-5=15\ m/s.$

Answer. $v=15$ $m/s$.