distance the cage:
1/2 .5 t^2
distance the rock:
1/2 9.8 (t-t0) where to is found
25=1/2 *.5t^2 or t= 10 check that or
distance the rock:
1/2 9.8 (t-10)
now set the distances equal, solve for time of impact t.
A cage descends a mine shaft with an acceleration of 0.5 m/s2. After the cage has travelled 25 metres,
a stone is dropped from the top of the shaft. Determine : 1. the time taken by the stone to hit the cage,
and 2. distance travelled by the cage before impact.
1 answer