What is the horizontal component of the velocity and the maximum height attained by the football?

December 22, 2009 by admin · 3 Comments
Filed under: Physics 
football
Andy H asked:

A player kicks a football at an angle of 30 degrees above the horizontal. The football has an initial velocity of 20.0 m/s.

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3 Responses to “What is the horizontal component of the velocity and the maximum height attained by the football?”
  1. anamika says:
    December 22, 2009 at 2:01 pm

    Horizontal component of velocity is( 20 cos30) = 17.32

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

  2. mis42n says:
    December 24, 2009 at 2:08 am

    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

  3. Trevor H says:
    December 27, 2009 at 8:40 am

    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

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