Since you mentioned 0.63 m/s, but the mass of the object is 100 g, there is uncertainty as to what unit is being used. You may want to include units in your future questions.
There are two stages of the motion of the object.
Stage 1: movement on the table, and
stage 2: free fall with horizontal velocity.
I will tackle stage 2 first, since all data are given.
First, consider vertical movement.
v0=0 m/s
a=-9.8 m/s²
Vertical distance,
Sv = v0*t + (1/2)at² = -1.5
Solving for t
t=sqrt(2Sv/a)
During this time, the object has travelled Sh metres.
Thus horizontal velocity, Vh
= Sh/t
For the stage 1 motions,
v0 = initial velocity, to be found
vh = final velocity, found above
acceleration = -μg due to friction
Distance, S=2.9 m
Use
vh²-v0²=2aS to solve for v0.
A popular pastime is to see who can push an object closest to the edge of a table without its going off. You push the 100 g object and release it 2.90 from the table edge. Unfortunately, you push a little too hard. The object slides across, sails off the edge, falls 1.50 to the floor, and lands 35.0 from the edge of the table. If the coeffecient of friction is .3 what is the objects speed as you released it?
I found the velocity in the x direction when the object falls off the table to be .63 m/s. how do i find the initial velocity?
2 answers
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