Question

At the top of a cliff 100 m high, Raoul throws a rock upward with velocity 15 m/s. How much later should he drop a second rock from rest so both rocks arrive simultaneously at the bottom of the cliff?
a. 5.05 s
b. 3.76 s
c. 2.67 s
d. 1.78 s

Answer 1

Answer:

d. 1.78s

Explanation:

The total time in the air for the second rock can be found with the next equation:

$h=v_{o}t+\frac{1}{2} gt^2$

where $h$ is the height, in this case 100m

$v_{o}$ the inicitial velocity wich is 0 since it came from rest

g is gravity and t is time

So we have:

$100=\frac{1}{2}(9.8m/s^2)t^2$

$t= \sqrt{\frac{200m}{9.8m/s^2} } =4.52s$

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

$t_{1}=2v_{o}/g =2(15m/s)/9.8m/s^2=3.06s$

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, $t_{2}$ is:

$h=v_{o}t+\frac{1}{2} gt^2$

$100=(15m/s)t_{2}+\frac{1}{2}(9.8m/s^2)t_{2}^2\\0=-100 + (15m/s)t_{2}+\frac{1}{2}(9.8m/s^2)t_{2}^2$

$0=-100 + (15m/s)t_{2}+(4.9m/s^2)t_{2}^2$

solving as a quadratic equation for time we get:

$t_{2}=3.24s$

So, the total time for the first rock is:

$t_{1}+t_{2}= 3.06s + 3.24s =6.3s$

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.