(25 %) Problem 2: An object rolls off a tabletop with a horizontal velocity v0x
ID: 1572988 • Letter: #
Question
(25 %) Problem 2: An object rolls off a tabletop with a horizontal velocity v0x 3.4 m/s. The table is at a height yo = 085 m, above the floor. Use a coordinate system with its origin on the floor directly beneath the point where the object rolls off the table, its horizontal x-axis lying directly beneath the object's trajectory, and its vertical y-axis pointing up 17% Part (a) How long, in seconds, is the object falling before it hits the floor? 17% Part (b) How far, in meters, does the object land from the edge of the tabletop? 17 % Part(c) What Is the vertical component of velocity in meters per second when the object hits the ground? Recall that the positive y- direction is upwards. = 17% Part (d) What is the magnitude of the velocity (it's speed) when it hits the floor? 17% Part (e) Find the angle of impact, in degrees below the horizontal. 17% Part (f) Enter an expression for the height of the object y in terms of x royo, and g. This expression should not depend on the timeExplanation / Answer
(a) Time = sqrt (2h/g) , h = .85 m, g = 9.8 m/sec^2 , time = .42 secs
(b) The horizontal distance travelled, x = v*t = 3.4 * .42 = 1.42 m
(c) The vertical component is given by: v = gt = 9.8 x .42 = .42
(d) The magnitude of speed : sqrt (vx^2 + vy^2) = 3.43 m/sec
(e) The angle is given by: arctan(vy/vx) = arctan(.42/3.4) = arctan(.124)
(f) The height , H = .85 - Vy^2 / 2 g
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