- #1
dizco29
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*4. 23) 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 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.
What I did:
Again I drew a diagram. I let my y=-40m from where my origin is. First I calculated the sound coming back to the soccer player, as that information was given in more detail. So I know the formula X=(Vi)(t) .
So (343)(3.00) = 1024m. So I know my hypoteneuse and my opposite side which is -40. So if I apply the pith. theorm, I get:
1024^2 - (-40)^2 = 1023.2184 m
So now I know my displacement.
At this point I want the total time from the kick to when it falls into the water so that I can calculate the initial velocity. I assume that I can use this formula:
xf = vyi t -1/2 gt^2
I assume that vyi is 0 because the soccer player kicks it straight and than it goes down (don't know if this is a valid assumption). Also, I subtract t-3 since we want to see when the rock hits the water, not includng the sound back to the the soccer player.
-40 = 0 - 1/2 (9.8)(t-3)^2
-40 = -4.9 (t^2-6t+9)
-40 = -4.9t^2 = 29.4t -44.1
= 4.9t^2 - 29.4t = 4.1
since I have a quadratic equation, I use a quadratic formula which gives me a time of 5.8 seconds. which is essentially time of flight. So when when I find the intial velocity, i plug it into this equation:
x = (v)(t)
1024 = (v) (5.8)
= 176 m/s
( I don't know, but I really doubt this answer for some reason, so if someone can correct it for me, it would be much appreciated!)
What I did:
Again I drew a diagram. I let my y=-40m from where my origin is. First I calculated the sound coming back to the soccer player, as that information was given in more detail. So I know the formula X=(Vi)(t) .
So (343)(3.00) = 1024m. So I know my hypoteneuse and my opposite side which is -40. So if I apply the pith. theorm, I get:
1024^2 - (-40)^2 = 1023.2184 m
So now I know my displacement.
At this point I want the total time from the kick to when it falls into the water so that I can calculate the initial velocity. I assume that I can use this formula:
xf = vyi t -1/2 gt^2
I assume that vyi is 0 because the soccer player kicks it straight and than it goes down (don't know if this is a valid assumption). Also, I subtract t-3 since we want to see when the rock hits the water, not includng the sound back to the the soccer player.
-40 = 0 - 1/2 (9.8)(t-3)^2
-40 = -4.9 (t^2-6t+9)
-40 = -4.9t^2 = 29.4t -44.1
= 4.9t^2 - 29.4t = 4.1
since I have a quadratic equation, I use a quadratic formula which gives me a time of 5.8 seconds. which is essentially time of flight. So when when I find the intial velocity, i plug it into this equation:
x = (v)(t)
1024 = (v) (5.8)
= 176 m/s
( I don't know, but I really doubt this answer for some reason, so if someone can correct it for me, it would be much appreciated!)