A stone is thrown upward from the edge of a cliff, reaches its maximum height, a

Question

A stone is thrown upward from the edge of a cliff, reaches its maximum height, and then falls down into the valley below. A motion diagram for this situation is given(Figure 1) , beginning the instant the stone leaves the thrower’s hand. Construct the corresponding motion graphs taking the acceleration due to gravity as exactly 10 m/s2 . Ignore air resistance. In all three motion graphs, the unit of time is in seconds and the unit of displacement is in meters. In plotting the points, round-off the coordinate values to the nearest integer.

1.Construct a graph corresponding to the stone's vertical displacement, y(t).

2.Construct a graph corresponding to the stone's vertical velocity,

vy(t).

3.Construct a graph corresponding to the stone's vertical acceleration,

ay(t).

You don't have to draw.

Just give me formula and key coordinate points so that I can draw please

Explanation / Answer

1.
For upward motion:
use:
y(t) = Vo*t +a*t^2
here a= -10m/s^2
Vo is initial velocity, velocity with which it is thrown
so,
y(t) = Vo*t -0.5*10*t^2
y(t) = Vo*t -5*t^2

Since it reaches max height at t=2 s, to find Vo use:
Vf = Vi + a*t
0 = Vo -10*2
Vo = 20 m/s

so,
y(t) = 20*t -5t^2

For downward motion:
y = yo-0.5*g*t^2

in temrs of y as a function of t
y(t) = y(2) - 0.5*10*t^2
y(t) = y(2) - 5*t^2

So to draw y(t) ve t use below equations:
for t between 0 to 2 s : y(t) = 20*t -5t^2
for t greater than 2 s and less than 6 s: y(t) = y(2) - 5*t^2

2)
Vy(t) = Vo-10*t : for all t between 0 to 6 s

3)
acceleration is always constant, which is -10 m/s^2 for all t
I have put negative sign as it is acting downward

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