1.A skateboarder shoots off a ramp with a velocity of6.8 m/s, directed at an ang
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Question
1.A skateboarder shoots off a ramp with a velocity of6.8 m/s, directed at an angle of63°above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis beparallel to the ground, the +y direction be verticallyupward, and take as the origin the point on the ground directlybelow the top of the ramp. 2. A criminal is escaping across a rooftop and runs offthe roof horizontally at a speed of 4.6m/s, hoping to land on the roof of an adjacent building. Airresistance is negligible. The horizontal distance between the twobuildings is D, and the roof of the adjacent building is2.0 m below the jumping-off point. Find the maximum value forD. 1.A skateboarder shoots off a ramp with a velocity of6.8 m/s, directed at an angle of63°above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis beparallel to the ground, the +y direction be verticallyupward, and take as the origin the point on the ground directlybelow the top of the ramp. 2. A criminal is escaping across a rooftop and runs offthe roof horizontally at a speed of 4.6m/s, hoping to land on the roof of an adjacent building. Airresistance is negligible. The horizontal distance between the twobuildings is D, and the roof of the adjacent building is2.0 m below the jumping-off point. Find the maximum value forD.Explanation / Answer
1.The distance between the point on the ground and the pointwhere the skateboarder lands is R = (u2 sin(2)/g) u = 6.8 m/s, = 63o and g = 9.8m/s2 2.The height of the first rooftop is H = (u2 * sin2/2g) u = 4.6 m/s, = 90o and g = 9.8m/s2 The height of the second rooftop is (H - 2.0) = (u2 *sin2(2')/g) solving the above equation for ' we get ' = (1/2) * sin-1[((H - 2.0) *g)1/2/u] -----------(1) Using trignometry we have tan' = (x/D) or D = (x/tan') x = 2.0 m and the value of ' is obtained from equation(1)Related Questions
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