Comments on: What is the horizontal component of the velocity and the maximum height attained by the football? http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/ smart sports discussion Mon, 21 Feb 2011 18:23:05 -0700 http://wordpress.org/?v=2.8.4 hourly 1 By: Trevor H http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/comment-page-1/#comment-3062 Trevor H Sun, 27 Dec 2009 14:40:36 +0000 http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/#comment-3062 Resolve the 20m/s velocity into vertical and horizontal components. This is easy for me to explain with a sketch, but I cannot do this. Draw a rectangle with the diagonal line drawn from the lower left corner to the upper right corner. This diagonal is marked 20 and represents the velocity and direction of the ball when kicked. The angle between this diagonal line and the bottom line of the rectangle is marked as 30°. Now the bottom line repesents the horizontal component of the velocity and the vertical line represents the vertical component of the velocity. Solve these two lengths by trig: Cv is calculated: Sin 30 = Cv ÷ 20 0.5 *20 = Cv Cv = 10m/s Ch is calculated: Cos 30 = Ch ÷ 20 0.866 * 20 = Ch Ch = 17.32m/s Maximum height reached by ball. Remember that at its maximum height the vertical velocity of the ball is zero. The horizotal velocity plays no part in this. V² = U² + 2 g s 0 = 10² + (2*9.8*s) 0 = 100 + 19.6s -100 = 19.6s s = -100÷ 19.6 s = -5.10m Note that the answer is negative because we took the acceleration due to gravity (g=9.8m/s² as positive downwards) The negative answer confirms that this is the height the ball will reach upwards. Answer: the ball will reach a height of 5.1m<a href="http://www.chefcookingware.com/967"> Trevor H</a> Resolve the 20m/s velocity into vertical and horizontal components. This is easy for me to explain with a sketch, but I cannot do this. Draw a rectangle with the diagonal line drawn from the lower left corner to the upper right corner. This diagonal is marked 20 and represents the velocity and direction of the ball when kicked. The angle between this diagonal line and the bottom line of the rectangle is marked as 30°. Now the bottom line repesents the horizontal component of the velocity and the vertical line represents the vertical component of the velocity. Solve these two lengths by
trig:
Cv is calculated:
Sin 30 = Cv ÷ 20
0.5 *20 = Cv
Cv = 10m/s

Ch is calculated:
Cos 30 = Ch ÷ 20
0.866 * 20 = Ch
Ch = 17.32m/s

Maximum height reached by ball.
Remember that at its maximum height the vertical velocity of the ball is zero. The horizotal velocity plays no part in this.
V² = U² + 2 g s
0 = 10² + (2*9.8*s)
0 = 100 + 19.6s
-100 = 19.6s
s = -100÷ 19.6
s = -5.10m
Note that the answer is negative because we took the acceleration due to gravity (g=9.8m/s² as positive downwards) The negative answer confirms that this is the height the ball will reach upwards.
Answer: the ball will reach a height of 5.1m Trevor H

]]> By: mis42n http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/comment-page-1/#comment-3061 mis42n Thu, 24 Dec 2009 08:08:11 +0000 http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/#comment-3061 Assume no air friction. vertical V sin(30)*20 = 10 horizontal V cos(30)*20 = 17.32 The ball starts at height 0 and ends up at height 0 s (distance travelled vertically in this case) = u(initial V)*t+1/2a(gravity)T^2 = 0 10*t+1/2(-9.8)T^2 = 0 10*t-4.9t^2 = 0 10t=4.9t^2 divide both sides by t 10=4.9t ; t=2.04 ball attains maximum height half way ie t=1.02 height is 10*1.02+1/2(-9.8)*1.02^2 10.2-5.09796=5.1m approx<a href="http://www.bigjobtools.com/srch/srch.php?q=sand+blasters"> mis42n</a> Assume no air friction.
vertical V sin(30)*20 = 10
horizontal V cos(30)*20 = 17.32
The ball starts at height 0 and ends up at height 0
s (distance travelled vertically in this case)
= u(initial V)*t+1/2a(gravity)T^2 = 0
10*t+1/2(-9.8)T^2 = 0
10*t-4.9t^2 = 0
10t=4.9t^2
divide both sides by t
10=4.9t ; t=2.04
ball attains maximum height half way ie t=1.02
height is 10*1.02+1/2(-9.8)*1.02^2
10.2-5.09796=5.1m approx mis42n

]]> By: anamika http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/comment-page-1/#comment-3060 anamika Tue, 22 Dec 2009 20:01:55 +0000 http://www.footballangst.com/blog/what-is-the-horizontal-component-of-the-velocity-and-the-maximum-height-attained-by-the-football/#comment-3060 Horizontal component of velocity is( 20 cos30) = 17.32 max height = ( 20^2 * (sin30)^2 ) \ 2g = 5 m (taking g = 10m\s)<a href="http://www.namingmybaby.com/muslim-baby-name.htm"> anamika</a> Horizontal component of velocity is( 20 cos30) = 17.32

max height = ( 20^2 * (sin30)^2 ) \ 2g
= 5 m (taking g = 10m\s) anamika

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