A soccer player kicks a rock horizontally off a 37.1 m high cliff into a pool of water. If the player hears the sound of the splash 3.98 s later, and the speed of sound is 343 m/s, what was the initial speed given to the rock?

1 Answer
Cameron G.
Nov 10, 2015

#153ms^-1#

Explanation:

Note: This answer may seem to be long and complicated, but it is actually only 4 major steps:

  1. Find the time the rock spends falling
  2. Find the distance from the splash to the kicker
  3. Find the horizontal distance the rock travelled
  4. Find the velocity of the rock

I have tried to keep it as simple as possible and explain the reasoning behind all the steps.

This answer does not take into account air resistance or any sort of friction at all.

First we have to calculate the time taken for the rock to hit the water. To do this we rearrange some common formulae:

#d = 1/2vt#

where #d = # the distance travelled
#v=# the final velocity before hitting the water
and #t = # the time taken to travel to the water

and

#v = at#

Where again #v =# the final velocity before hitting the water
#a =# the acceleration due to gravity (#9.8ms^-1#)
and again #t = # the time taken to travel to the water

Substituting the first into the second we get:
#d = 1/2at^2#
Which we can rearrange to get:
#t = root2 ((2d)/a)#
Substituting the known values into the equation we get:
#t = root2((2 xx 37.1m)/(9.8ms^-1)) = 2.751622898s#

So we have the time taken for the rock to hit the water.

Next we need to find the time taken for the sound of the splash to reach the kicker.
#3.98 - 2.751622898 = 1.228377102s#
Then we use the speed of sound to calculate the distance from the splash to the kicker.
#343 xx 1.228377102 = 421.3333461m#

We now need to know the horizontal distance the rock travelled. We use Pythagoras' Theorem to do this:

enter image source here

#x = root2(421.3333461^2 - 37.1^2) = 419.6967697m#

We now know the horizontal distance travelled and time taken before the rock hits the water, and it is simple to find out the velocity from this:

#v = d/t = (419.6967697m)/(2.751622898s) = 152.52669942ms^-1#

# = 153ms^-1# (The question gave values accurate to 3 s.)

This seems to be quite a high value and if it is converted to km/hour it comes out as a staggering:

#152.52669942 xx (60xx60)/1000 = 549.0971791kmh^-1#

This very high speed seems to be almost superhuman as the fastest ever recorded football kick was 210 km/hour (from this site) and the world record for the fastest golf drive is 349.38km/h.

This makes me wonder if I have missed something in my working or if I have got one of the formulas wrong.

Then again it could just be that it is a made-up question and the values were not thought through.

Related questions
See all questions in Projectile Motion
Impact of this question
17542 views around the world
You can reuse this answer
Creative Commons License