SSenn 2D Deg (4%) Problem 18. An object rolls off a tabletop with a horizontal v
ID: 1882454 • Letter: S
Question
SSenn 2D Deg (4%) Problem 18. An object rolls off a tabletop with a horizontal velocity Ox-34 ms. The table is at a heightyo = 0.35 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? 1 7% 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 v-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 ofx vx yo and g This expression should not depend on the time Grade Summary Potential 100 789 Submissions Attempts remaining:2 (0% per attempt) detailed view 0 END VOx BACKSPACE CLEAR Submit Hint I give up!Explanation / Answer
[a] since the ball rolls off the table it starts with a vertical velocity of 0.
So distance it falls versus time is given by:
s = (1/2)gt^2
t^2 = 2*s/g
t = SQRT(2*s/g) = SQRT(2*0.35/9.8)
t = 0.267 seconds
(b)The horizontal or x_displacement = (Vox)•t
= (3.4)•(0.267)
= 0.91 m
[c] The ball is falling for 0.267 sec under the constant acceleration of gravity.
v = g*t = 9.8*(0.267)
v = 2.62 m/s
[d] the ball travels a horizontal distance of 3 m in the time it takes to hit the floor. the horizontal velocity is just distance over time so:
v = d/t = 0.91/0.267
v = 3.41 m/s
[d]
= tan ¹ [ (2.62) (3.41) ] = 37.53°
Hope this helps you.
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