Point A is located at (-3, -5)
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Answer:
6. The distance Rampy Ronald would have travelled horizontally is approximately 48.428 feet for an actual maximum height of 21.486 feet
7. Rampy Ronald is going to make the jump
Please find attached the graph of the motion of Rampy Ronald, created with Microsoft Excel
Step-by-step explanation:
6. The given parameters are;
The width of the ravine = 80 ft
The inclination of the ramp with which Rampy Ronald will jump = 30°
The distance from the edge of the ravine the ramp ends = 5 feet
The speed with which Rampy Ronald will make the jump = 60 mph
The maximum height he will reach = 30 feet
Therefore, we have;
The height of the end of the ramp = 15 × sin(30°) = 7.5
The height of the end of the ramp = 7.5 feet
The height of the jump = 30 - 7.5 = 22.5
The vertical velocity = 60 × sin(30°) = 30
The vertical velocity = 30 mph
The horizontal velocity = 60 × cos(30°) = 30·√3
The horizontal velocity = 30·√3 mph
From the kinematic equation, v² = u² - 2gh, we have;
u² = 2gh
h = u²/2g = 900/(2 × 32.17405) ≈ 13.986
The maximum height reached 13.986 + 7.5 = 21.486
From v = u - gt
t = u/g = 30/32.17405 ≈ 0.932
t ≈ 0.932 s
The distance Rampy Ronald would have travelled horizontally = 30·√3 × 0.932 ≈ 48.428
The distance Rampy Ronald would have travelled horizontally ≈ 48.428 feet
7. The time before Rampy Ronald reaches ground level again, is given by the following relation;
h = 1/2·g·t²
21.846 = 1/2 × 32.17405 × t²
t = √(21.846/16.087025) ≈ 1.165
t ≈ 1.366 s
The horizontal distance traveled is therefore;
The horizontal distance traveled from maximum height is 30·√3 × 1.165 ≈ 60.535
The horizontal distance traveled from maximum height is ≈ 70.979 feet
The total horizontal distance traveled ≈ 48.428 + 60.535 = 108.963
The total horizontal distance traveled ≈ 108.963 feet
The total horizontal distance required to make it across the ravine = 5 feet + 80 feet = 85 feet
Therefore, Rampy Ronald is going to make the jump
