Answer:
h=17m
v=18.6 m/s
Explanation: The question can be solved by applying kinematic equations of motion
Data
u=0
a=g
t=1.9 secs
firstly to calculate the height
[tex]s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m[/tex]
to find the final velocity
[tex]v=u+at\\v=0+9.81*1.9\\v=18.639[/tex]
The acceleration graph is straight line of equation y=9.8 as acceleration is constant:
Velocity graph is given by y=9.8x ( y as velocity and x as time):
Displacement graph is given by y=4.9x^2 ( x as time, y as displacement):
These graphs are only applicable from x=0 to x=1.9 ... ignore the other graph sections
