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1. ### 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

2. ### A quarterback takes the ball from the line of scrimmage, runs backwards for 10.0 yards, then runs sideways par?

the displacement would simply be the hypotenuse of the right triangle formed by 10, and 15. There fore the displacement or d= sq route of 10 sq + 15 sq or 100+225... The square route of 325 is (18.027756), according to my phone calculator at least. This problem is really very simpley...

2 Answers · Science & Mathematics · 27/08/2011

3. ### 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

4. ### 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

u have to use the range formula R=(u^2 sin2(angle))/g u-intial velocity-28m/s This gives the total distance traversed.Now find the distance traversed in 2.9s and subtract the two distances. plug this in s=1/2 *g*t^2 and u should get the answer.

3 Answers · Science & Mathematics · 29/01/2008

5. ### A quarterback claims that he can throw the football a horizontal distance of 169.1 m (185 yd). Furthermore, he?

range = v0^2 sin(2 theta) /g 169.1 = v^2* sin (2*25)/9.9 169.1 = v^2 (sin 50) /9.8 169.1 = v^2 * 0.766/9.8 v^2 = 169.1 / 0.0782 v^2 = 2162.4 v = 46.50 m/s

2 Answers · Science & Mathematics · 13/12/2016

6. ### A quarterback claims that he can throw the football a horizontal distance of 169.1 m (185 yd). Furthermore, he?

range = v0^2 sin(2 theta) /g 169.1 = v^2* sin (2*25)/9.9 169.1 = v^2 (sin 50) /9.8 169.1 = v^2 * 0.766/9.8 v^2 = 169.1 / 0.0782 v^2 = 2162.4 v = 46.50 m/s

3 Answers · Science & Mathematics · 28/01/2011

7. ### Please Help! A quarterback claims that he can throw the football a horizontal distance of 196.6 m (215 yd). Fu?

Height at any time t = y(t) = h + Uy t - 1/2 g t^2 Distance at any time t = x(t) = Ux t g = 9.807 m/s^2 (ft/s^2) Launch height above ground H = 2.000 meters (feet) Impact (target) elevation y(T) = 2.000 m (ft...

1 Answers · Science & Mathematics · 20/09/2012

8. ### 2D Motion Kinematics?

1. A quarterback takes the ball from the line of scrimmage, runs backward for 3 yards...

1 Answers · Science & Mathematics · 05/10/2019

9. ### A quarterback can throw a receiver a high, lazy "lob" pass or a low, quick "bullet" pass.?

a] Flight time will give you V0y, by d = V0yt + .5at^2, where d = 0 Since you know far it went in how much time you know V0x. Launch angle = arctan{V0y/V0x] b,c] I think here you have to use the V0x and V0y...

2 Answers · Science & Mathematics · 03/11/2012

10. ### In a game of American football, a quarterback takes the ball from the line of scrimmage, runs backward for 10.?

This is a trig, not physics, problem. The resultant displacement, measured from where the ball lay before it was hiked, is D^2 = X^2 + Y^2 and theta = arcsin(Y/D) is the angle measured...

2 Answers · Science & Mathematics · 05/04/2011