A football is kicked straight up into the air; it hits the ground 4.0s later. How do I find the greatest height and the speed it leaves the kicker's foot with no other info? I'm sure it's probably some really easy physics concept, but it's eluding me. First answer that helps gets best answer. Thanks!
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t = Sqrt( 2(h)/ g )
if the total time is 4 seconds, then half that is the point at which its at its peak
so your t= 4/2 = 2
g is gravity: 9.81m/s/s
So now to set it up with the above equation:
2=sqrt(2(h)/9.81)
-square both sides to remove the square root:
4=2(h)/9.81
-multiply both sides by 9.81 and then divide both side by 2 leaves you with:
19.62meters=h
AND now that you know the height you can now find your initial velocity, WOO!
so we can plug it into the max height equation:
Ymax=[ Vi^2 * sin^2(theta) ]/ 2*g
Ymax=19.62m=[ Vi^2 * sin^2( 90 ) ]/ 19.62
-multiply the Ymax by the 2g (which is coincidentally 19.62^2) and calculate the sin^2(90)
384.94m= Vi^2 * 1
-and finally multiply out the 1 and sqrt both sides
19.62m/s= Vi
-which is also coincidentally hour height, what a strange problem hahaha
I hope this made sence, since i really like best answer :D
if the total time is 4 seconds, then half that is the point at which its at its peak
so your t= 4/2 = 2
g is gravity: 9.81m/s/s
So now to set it up with the above equation:
2=sqrt(2(h)/9.81)
-square both sides to remove the square root:
4=2(h)/9.81
-multiply both sides by 9.81 and then divide both side by 2 leaves you with:
19.62meters=h
AND now that you know the height you can now find your initial velocity, WOO!
so we can plug it into the max height equation:
Ymax=[ Vi^2 * sin^2(theta) ]/ 2*g
Ymax=19.62m=[ Vi^2 * sin^2( 90 ) ]/ 19.62
-multiply the Ymax by the 2g (which is coincidentally 19.62^2) and calculate the sin^2(90)
384.94m= Vi^2 * 1
-and finally multiply out the 1 and sqrt both sides
19.62m/s= Vi
-which is also coincidentally hour height, what a strange problem hahaha
I hope this made sence, since i really like best answer :D