This is approximately 6.9 m/s. Let’s determine the vertical and horizontal components of this velocity.
Vertical = √(112 * sin 25) * sin 25
This is approximately 2.91m/s. The direction of this velocity is downward.
Horizontal = √(112 * sin 25) * cos 25
This is approximately 6.24 m/s.
As the ball falls 5 meters, its vertical velocity increases at the rate of 9.8 m/s each second. Let’s use the following equation to determine the time to fall 5 meters.
d = vi * t + ½ * g * t^2
5 = √(112 * sin 25) * sin 25 * t + ½ * 9.8 * t^2
4.9 * t^2 + √(112 * sin 25) * sin 25 * t – 5 = 0
t = ([-√(112 * sin 25) * sin 25 ± √[√(112 * sin 25) * sin 25)^2 – 4 * 4.9 * -5]) ÷ 9.8
± √(8.4501085 + 98) = ±√106.4540109
This is approximately ± 10.32 m/s.
t = ([-√(112 * sin 25) * sin 25 ±√106.4540109] ÷ 9.8
t = ([-√(112 * sin 25) * sin 25 +√106.4540109] ÷ 9.8 = 1.34951362
This is approximately 1.35 seconds. To determine the answer to this question, multiply the horizontal component of the sphere’s velocity at the bottom of roof by the time.
d = √(112 * sin 25) * cos 25 * ([-√(112 * sin 25) * sin 25 +√106.4540109] ÷ 9.8
This is approximately 11.4 meters
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oldprof say: No no Phil, that was the Pegasus that it morphed into and then flew away. The cylinder actually hit tera firma. And here's where:
Total energy TE = PE = mgS sin(theta) above the eave of the roof. And that equals TE = KE = 1/2 mv^2(1 + k) at the eave as the potential energy is converted into kinetic. The TE is fixed so we have mgS sin(theta) = 1/2 mv^2(1 + k) and find v = sqrt(2gS sin(theta)/(1 + k)) and then Vx = v cos(-theta) and Vy = v sin(-theta).
We solve the quadratic y(t) = 0 = h - v sin(-theta) t - 1/2 g t^2 to find t the trajectory time of fall between the eave and the ground. And with that X = v cos(-theta) t = ? the standoff distance from the eave you are looking for.
So h = 5 m, v = sqrt(2*9.8*8*sin(radians(25))/(1 + 2/5)) = 6.88 m/s off the edge (eave), so we have Vx = 6.88*cos(radians(25)) = 6.23 m/s and Vy = 6.88*sin(radians(-25)) = -2.91 m/s. So the quad is:
y(t) = 0 = 5 - 2.91t - 4.9 t^2; which has the solution t = .756 seconds. And in that time the ball turned cylinder goes X = 6.23*.756 = 4.71 meters. ANS.
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Philomel say: It doesn't hit the ground. When it changed from a sphere to a cylinder it became weightless and flew away.
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