Question

An inquisitive physics student and mountain climber
climbs a 50.0-m-high cliff that overhangs a calm pool of
water. He throws two stones vertically downward, 1.00 s apart,
and observes that they cause a single splash. The first stone
has an initial speed of 2.00 m/s. (a) How long after release
of the first stone do the two stones hit the water? (b) What
initial velocity must the second stone have if the two stones
are to hit the water simultaneously? (c) What is the speed of
each stone at the instant the two stones hit the water?

Answer 1

Firstly, you need to figure out how long it takes stone 1 to reach the water. Use the formula:\[\Delta y=u_y+\frac{1}{2}at^2\]You know \(\Delta y\) (50.0 m), \(u_y\) (2.00 m/s), and \(a\) (9.81 m/s/s). What's t?

Answer 2

Sorry, that formula is \[\Delta y=u_yt+\frac{1}{2}at^2\]

Answer 3

3s

Answer 4

The second stone was released 1 second after the first stone, and hit the water at the same time, so the total travel time is 2 seconds. Now, using the same formula as before:\[\Delta y=u_{y2}t+\frac{1}{2}at^2\]This time, your unknown is \(u_{y2}\). You have 't', I just told you what that was :)

Answer 5

Initially that is what I did. However, my answer differs from the books by 0.1 sec. Since i am learning and do not feel confident. i needed to ask

Answer 6

For a difference of that size, it is possibly a rounding error.

Answer 7

For example, they might have used 9.8, or 10 as the acceleration due to gravity

Answer 8

thanks for the assistance