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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

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

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

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

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

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

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

1. A

**quarterback**takes the ball from the line of scrimmage, runs backward for 3 yards...1 Answers · Science & Mathematics · 05/10/2019

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

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

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