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1. ### A quarterback threw a football at 28 m/s at a 45° angle. If it took the ball 2.9 s to reach the top of its pat

The easy answer is twice 2.9 seconds,or 5.8 seconds, since time up usually = time down. But if you consider that it might land on the ground, it would be in the air a little longer, since the falling distance would be larger than the rising distance.

2 Answers · Science & Mathematics · 24/08/2009

2. ### A quarterback throws a football at 21 m/s at a certain angle above the horizontal. If it took the ball 2.6 s t?

It would be 2 times the time it took the ball to reach its peak. In this case, it would be 5.2

3 Answers · Science & Mathematics · 04/11/2009

3. ### A quarterback throws a football at 25 m/s at a certain angle above the horizontal. If it took the ball 2.5 s t?

5 seconds. When at the top of its path it's halfway through its total trajectory. Think about the parabolic graph and that'll make sense. That means to find out how long it was in the air, you just have to multiply the halfway point by 2.

2 Answers · Science & Mathematics · 25/01/2010

4. ### Physics Help: determine the speed with which this quarterback must throw the ball.?

By R = u^2sin2θ/g =>164 = u^2sin(2 x 30.8)*/9.8 =>u = √1827.09 =>u = 42.74 m/s

2 Answers · Science & Mathematics · 24/09/2010

5. ### PHYSICS HELP!!! determine the speed with which this quarterback must throw the ball.?

the highest point of the arc of his throw is at a distance of 84.55 meters at a height x/84.55= tangent 30 degrees X= 84.55(.5774 ) = 48.82 meters V^2 = 2AS V^2 = 2(9.81)(48.82) = 957.85 V = 30.95 m/s

3 Answers · Science & Mathematics · 14/11/2008

6. ### A rookie quarterback throws a football with an initial upward velocity component of 16.3m/s and a horizontal?

Given Ux = ucosθ = 20.6 m/ & Uy = usinθ = 16.3 m/s (1) By t = usinθ/g = 16.3/9.8 = 1.66 sec (2) By H = (usinθ)^2/2g = (16.3)^2/(2 x 9.8) = 13.56 m (3) By T = 2t = 3.32 sec (4) By R = [Ux] x T =>R = 20.6 x 3.32 = 68.39 m

2 Answers · Science & Mathematics · 19/07/2013

7. ### A quarterback takes the ball from the line of scrimmage, runs backward for 8 yards, and then runs sideways par?

48 yards.

2 Answers · Science & Mathematics · 26/09/2011

8. ### A football quarterback is moving straight backward at a speed of 3.00 m/s when he throws a pass to a player 16?

Symbols used: v→ = horizontal velocity R = range (horizontal distance the ball moves) v = actual velocity u↑ = initial vertical velocity component v↑ = final vertical velocity component t = time for ball to reach maximum height T = time in air h = height ball reaches above point of release...

2 Answers · Science & Mathematics · 03/10/2012

9. ### A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 17?

as g=9.8m/s^2 d=d0+vt = 17+2t d=vt = v0cos(25)t v0cos(25)t= 17+2t ...(1) h=vyt+1/2gt^2 0= v0sin(25)t-4.9t^2 v0sin(25)t= 4.9t^2 ...(2) (2)/(1) tan(25)= 4.9t^2/(17+2t) 0.4663= 4.9t^2/(17+2t) 4.9t^2+0.9326t+7.9272=0 t= 1.3706s = 1.37[s] (Flight duration) v0cos(25)t=17+2t v0cos(25)*1.3706=17+2*1.3706 v0cos(25)= 14.4033= 14.4[m/s] (horizontal velocity) v0= 15.8923= 15...

2 Answers · Science & Mathematics · 05/03/2014

10. ### football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m s?

1st Average velocity = 15/2.5 = 6 m/s 2nd Average velocity = -3/1.75 ≈ -1.714 m/s 3rd Average velocity = 21/5.2 ≈ 4.038 m/s Total average velocity = total displacement/ total time Total displacement = 15 – 3 + 21 = 33 m Total time = 2.5 + 1.75 + 5.2...

2 Answers · Science & Mathematics · 09/09/2013

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