Answer:
d. 1.78s
Explanation:
The total time in the air for the second rock can be found with the next equation:
[tex]h=v_{o}t+\frac{1}{2} gt^2[/tex]
where [tex]h[/tex] is the height, in this case 100m
[tex]v_{o}[/tex] the inicitial velocity wich is 0 since it came from rest
g is gravity and t is time
So we have:
[tex]100=\frac{1}{2}(9.8m/s^2)t^2[/tex]
[tex]t= \sqrt{\frac{200m}{9.8m/s^2} } =4.52s[/tex]
For the fist rock we need to find the time it takes to go up and go back down to the height it was launched:
that time is
[tex]t_{1}=2v_{o}/g =2(15m/s)/9.8m/s^2=3.06s[/tex]
and the time the fist rock is going down from that point, we can find in a similar way we did for the fist rock, [tex]t_{2}[/tex] is:
[tex]h=v_{o}t+\frac{1}{2} gt^2[/tex]
[tex]100=(15m/s)t_{2}+\frac{1}{2}(9.8m/s^2)t_{2}^2\\0=-100 + (15m/s)t_{2}+\frac{1}{2}(9.8m/s^2)t_{2}^2[/tex]
[tex]0=-100 + (15m/s)t_{2}+(4.9m/s^2)t_{2}^2[/tex]
solving as a quadratic equation for time we get:
[tex]t_{2}=3.24s[/tex]
So, the total time for the first rock is:
[tex]t_{1}+t_{2}= 3.06s + 3.24s =6.3s[/tex]
This means that the second rock must be dropped 6.3s - 4.52 s = 1.78 seconds later, wich is the difference in the times that it takes for each rock to get to the bottom if the cliff.
