#$&*

course Phy121

3.21.11 at 9:00pm

RandomAssignment8#$&*

course Phy121

3.14.11 at 6:45pm

A projectile leaves the edge of a table and, while traveling horizontally at a constant 37

cm/s, falls freely a distance of 149 cm to the floor. If its vertical acceleration is 980

cm/s2, how long does it take to fall and how far does it travel in the horizontal direction

during the fall?'ds=149cm

a=980cm/s^2

v0=0

vf=37cm/sec

@& 37 cm/s is a horizontal velocity, not vertical.

Your vertical velocity is initially zero, since the ball is moving only in the horizontal

direciton when it leaves the tabletop.

So your vertical motion is characterized by `ds = 149 cm, initial velocity is 0 and

acceleration is 9.8 m/s^2, or 980 cm/s^2.

What can you therefore determine about the vertical motion, using the equations of motion?*@

&&&&&&&&&&&&&&&& vf^2 = v0^2 + 2a'ds

=2(980cm/sec^2)(149cm)

=292040

vf =540.407cm/sec&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

vAve= 18.5cm/sec

'ds=v0'dt+0.5a'dt^2

'ds=0.5a'dt^2

'dt=sqrt(2'ds/a)

'dt=sqrt((2*149cm)/(980cm/sec^2)

'dt=0.55

'ds=0.55sec*18.5cm/sec

@& You need to analyze the horizontal motion.

The time interval is the same as for the vertical motion. Your analysis of the vertical

motion should have given you the time interval.

The initial velocity in the horizontal direction is 37 cm/s. While falling there is no force

in the horizontal direction, therefore zero acceleration in this direction.

How far does the ball therefore travel in the horizontal direciton, during its fall?*@

&&&&&&&&&&&& a=0, 'dt=0.55sec, v0=37cm/sec If acceleration is zero, v0=vf.

'ds=(V0+vf)/2 *'dt

=74cm/sec/2*0.55sec