How Do You Calculate the Initial Speed of a Rock Kicked Off a Cliff?

[Melchior25]

Homework Statement

A soccer player kicks a rock horizontally off a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.07 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

The Attempt at a Solution

So here is what I did. I first figured out the time it takes for the rock to hit the water by using the equation.

\frac{1}{2}gt^2 = 40

And I got 2.85714 s. From there I subtracted that number from the original time of 3.07 s to get .21286 s, which is the time for speed of sound to reach the top again. I them found the hyp distance by taking (.21286)(343) = 73.011 m

After that I drew a triangle with 73.011 m as the hyp. But now I'm stuck. I don't know how to still get the initial speed of the rock. I'm not even sure what equation I should use.

Please Help.

 

[Avodyne]

You know the hyp (73.011 m), and you know the vertical side (40 m). So you can compute the horizontal side. You also know the time in the air (2.857 s). Do you know a formula that relates horizontal distance to initial horizontal velocity and time?

(By the way, I did not check any of your numbers, but they look reasonable.)

 

[Melchior25]

Sorry, I forgot to mention that, but yes I did figure out the the x distance to be 61.0787 m. I also figured out all the angles inside the triangle because it forms a right triangle.

The equation I used was yf = (vyi)(t) + (1/2)(Ay)(t^2)

I used 2.857 s as my time and set yf = to -40. I then solved for my velocity but I got .000033 which I know is definitely not right.