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  1. ...30° t = 6 s So the time to maximum height is: t / 2 = 6 / 2 = 3 sec Solve for Voy (vertical component of velocity) t = Voy /g...

    1 Answers · Science & Mathematics · 24/09/2013

  2. By F = ∆P/∆t =>F = m∆v/∆t =>F = [0.57 x 35]/0.016 =>F = 23121.88 N

    2 Answers · Science & Mathematics · 03/01/2012

  3. a=∆v/∆t =(0 - 2.4 yds/s)/(0.04 min * 60 s/min) = -1 yd/s^2

    1 Answers · Science & Mathematics · 06/03/2008

  4. And???? What do you want to know? The range is 2 . V^2 , cos ø . sin ø / g = 2 . 48^2 cos 37 . sin 37 / 32 = 69 ft

    2 Answers · Science & Mathematics · 23/03/2011

  5. c) based on zero air resistence and uniform gravity, the velocity when it hits the ground is 20.0m/s (no friction to stop or slow it). The rest of the questions are based on vectors and trig function, but I can't remember them

    1 Answers · Science & Mathematics · 15/03/2007

  6. A football game begins with a kickoff in which the ball travels a horizontal... are in metres....gravitational acceleration g = 9.806 m/ sec ^2 45 yards = 45*0.9144 = 41.148 Θ angle = 39° ; sin 39...

    4 Answers · Science & Mathematics · 19/10/2014

  7. ... Vsin(theta) 20 sin 37 = 12 m/s 12m/s / 9.8 m/s^2 = 1.23 secs vertical travel To find distance use D= Vt + .5AT^2...

    2 Answers · Science & Mathematics · 12/12/2006

  8. ...48 m/s Time football in the air: t = 2 v/a = 2 x 14.34 / 9.8 = 2.93 sec distance the football to be catch: s = t x v = 2.93 x 20.48 = 60.01 m The receiver is...

    1 Answers · Science & Mathematics · 10/06/2012

  9. ... = initial velocity = 27 (given) g = acceleration due to gravity = 9.8m/ sec ^2 (constant) s = maximum height attained by the football Substituting values, 0 - (27 * sin 31)^2 = 2(-9.8)s ...

    7 Answers · Science & Mathematics · 15/05/2008

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