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rockcandy

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Member advised to place separate questions in separate threads (One question per thread)

## Homework Statement

1. A diver jumps from a height of 10m leaving with a velocity of 8m/s find the time for which the diver is in the air and the speed at which the diver enters the water?

2. A ball is launched from point A on a horizontal plane with an angle of 55*. It passes two walls both 2m high. Wall one is 2m from point A. Find the speed of the ball and the Wall two distance from point A

The attempt at a solution

The attempt at a solution

1. V

_{y}= 8m/s d

_{y}= 10m a

_{y}=-9.8m/s

^{2}

V

_{fy}

^{2}= V

_{iy}

^{2}+ 2a

_{yd Vy=√(8m/s)2 + 2(-9.8m/s2)(-10m) Vy = 16.1 m/s -10 m = 8 m/s t + 1/2 (-9.8m/s2) t2 0 = -4.9 m/s2 t2 + 8m/s t + 10m t = 2.45 s or - 0.82 s 2. Vx = VicosΘ and Vy = VisinΘ h =2m and horizontal distance = 2m dy= Viyt + 1/2 at2 dx = Vixt + 1/2at2 since horizontal acceleration = 0 ; dx = Vixt dy= Viyt + 1/2 at2 dx = Vixt ; 2 m = VicosΘ t 2 = VisinΘ t - 1/2 (9.8m/s2) t 2 2 = VisinΘ (2m /VicosΘ) - 1/2 (9.8m/s2)(2m /VicosΘ)2 2 = 2m tan Θ - 4.9m/s2 (12.15 m 2 /Vi2 ) 2 = 2.85 m - 59.535 / Vi2 Vi = √70.04 Vi = 8.36 m/s Im stuck on how to get the distance assuming this is correct. If I use the max range formula x = Vi2 sin2Θ / g x = (8.36m/s)2 sin 2(55) / 9.8 m/s2 x = 6.69 m If I take half this distance ( 6.69m)(.5) = 3.34 m I know that it is reaching a height of 2 m twice, and given that the parabola is symmetrical I know that on the opposite side it is also 2m away from Wall 2. 3.34 m - 2 m = 1.34 m , which the remaining distance between the wall to the half the max range. therefore 1.34m x 2 = 2.68m which is the total distance between Wall 1 and Wall 2 So to find the distance from Wall 2 to Point A just add the total = 2.68m + 2 m = 4.68m }

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