Asked by Carissa
A catapult launches a stone at an angle of 20° above the horizontal at a speed of 26.8 m/s. Assuming the stone begins its trajectory at the origin, determine the distance in meters from the origin that the stone achieves after 1.05 s.
Note: The stone will change position in the x and y direction, so you need to calculate the straight-line distance between the origin and the position of the stone.
Note: The stone will change position in the x and y direction, so you need to calculate the straight-line distance between the origin and the position of the stone.
Answers
Answered by
Henry
Vo = 28.6 m/s[20o]
Xo = 28.6*Cos20 = 25.18 m/s.
Yo = 28.6*sin20 = 9.78 m/s.
Dx = Xo * t = 25.18m/s * 1.05s. = 26.4 m
Xo = 28.6*Cos20 = 25.18 m/s.
Yo = 28.6*sin20 = 9.78 m/s.
Dx = Xo * t = 25.18m/s * 1.05s. = 26.4 m
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