Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J