Aball is thrown horizontally from the top of a building 14.9 m high. the ball strikes the ground at a point 107 m from the base of the building. the acceleration of gravity is 9.8 m/s 2 . find the time the ball is in motion. find the initial velocity of the ball. answer in units of m/sfind the x component of its velocity just be- fore it strikes the ground. answer in units of m/sfind the y component of its velocity just be- fore it strikes the ground. answer in units of m/s

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

a straight line sloped

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.