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  1. The ball will have a horizontal range given by .. R = u².sin2θ / g .. standard equ.of motion R = 17.0².sin62 / 9.80 = 26.0 m Time of flight t = 2u.sinθ / g .. another standard equ. of motion t = 2 x 17.sin31 / 9.80 .. .. t = 1.79s So the receiver has to run (26.0...

    2 Answers · Science & Mathematics · 04/11/2011

  2. Use the range equation [ R = u².sin2θ / g ] to find the throwing angle θ (to horizontal) .. sin 2θ = Rg / u² .. .. 30m x 9.80 / (20m/s)² .. .. 0.7350 .. .. 2θ = 47.30º .. .. θ = 23.65º Max. vertical height, h obtained from [ Vv²...

    3 Answers · Science & Mathematics · 15/10/2011

  3. That is one messed up play.

    2 Answers · Science & Mathematics · 09/10/2011

  4. dx = distance the ball was thrown dx = Vix(t) dx = (Vicos35deg)t dx = (25 m/scos 35 deg)(2.55s) dx = 52.22 m answer Hope this helps. teddy by

    2 Answers · Science & Mathematics · 01/10/2008

  5. Let us consider first just the horizontal component of the ball's velocity, h. Let's say the time the ball is in the air is t. The receiver runs for t+3 seconds and covers a distance of 7.5 * (3+t). The ball must thus travel a distance of 3.5...

    2 Answers · Science & Mathematics · 15/04/2010

  6. (a)♣ the vertical component of ball’s velocity is Vy=Voy –g*t; at vertex t=Tc/2 and Vy=0= Voy –g*(Tc/2), hence Voy=0.5g*Tc; (b)♠ the R-guy catches and his position is x=D+Vr*Tc; horizontal...

    1 Answers · Science & Mathematics · 14/10/2008

  7. I guess the net change is 9 yards sideways and 54 - 15 = 39 yards downfield. So, the football has moved √(39² + 9²) = √(1521 + 81) = √1602, about 40 yards.

    2 Answers · Science & Mathematics · 02/10/2007

  8. Maximum range is achieved if the ball is thrown at 45 degrees to horizontal. Initial vertical V component = (sin 45 x 27) = 19.1m/sec. Time to max. height = (v/g) = 19.1/9.8 = 1.949 secs. b) Double it, = 3.898 secs. time...

    2 Answers · Science & Mathematics · 16/09/2014

  9. Without a horizontal velocity component the ball can only be thrown vertically upwards. I suppose it could be thrown high enough for the runner to get to where the thrower is in time to catch the vertically falling ball (if you see what I mean !) .. but that's not what I would...

    2 Answers · Science & Mathematics · 10/11/2011

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