Question

A stone is dropped from the top of a cliff 120 metres high. After one second, another stone is thrown

down and strikes the first stone when it has just reached the foot of the cliff. Find the velocity with

which the second stone was thrown.

Answer 1

Solution.

$h_1=120m;$

$v_{01}=0;$

$\Delta t=1s;$

$h_2=120m;$

$t2=t1-\Delta t;$

$v_{02}-?$;

$h_1=v_{01}t_1+\dfrac{gt_1^2}{2};$ $v_{01}=0, h_1=\dfrac{gt_1^2}{2}\implies t_1=\sqrt{\dfrac{2h_1}{g}};$

$t_1=\sqrt{\dfrac{2\sdot120m}{9.81m/s^2}}=4.95s;$

$t_2=4.95s-1s=3.95s;$

$h_2=v_{02}t_2+\dfrac{gt_2^2}{2};$

$v_{02}t_2=h_2-\dfrac{gt_2^2}{2};$

$v_{02}=\dfrac{h_2}{t_2}-\dfrac{gt_2}{2};$

$v_{02}=\dfrac{120m}{3.95s}-\dfrac{9.81m/s^2\sdot3.95s}{2}=11m/s;$

Answer: $v_{02}=11m/s.$