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We now can use the ‘classical’ mechanics formulae for constant acceleration motion.

S(t) = S(0) + V(0)*t + 0.5*a*t^2 (1)

and,

V(t) = V(0) + a*t (2)

t : time since being thrown in seconds

a: acceleration

a = g = -9.81 m/s^2 ; negative because it points downward

S(t) : position (vertical) at time t in meters;

S(0) = 80m; starting position

V(t) : velocity at time t

V(0) = 15 m/s

Take t’ to be the time that the stone hits the ground, take equation (1) and fill in all the data you have:

S(t’) = 0 = 80 + 15*t’ -0.5*9.81*(t’)^2

Solve this quadratic equation for t.

You’ll get two solutions, one of which is negative time; discard that one.

Fill in your solution for t’ into the velocity equation (2) and get the velocity as well.

[tex]\tiny\sf{Hope \: this \: will \: help \: you \: \: \: \: \: \: \: \: \: \: \: }[/tex]

Answer:

We now can use the ‘classical’ mechanics formulae for constant acceleration motion.

S(t) = S(0) + V(0)*t + 0.5*a*t^2 (1)

and,

V(t) = V(0) + a*t (2)

t : time since being thrown in seconds

a: acceleration

a = g = -9.81 m/s^2 ; negative because it points downward

S(t) : position (vertical) at time t in meters;

S(0) = 80m; starting position

V(t) : velocity at time t

V(0) = 15 m/s

Take t’ to be the time that the stone hits the ground, take equation (1) and fill in all the data you have:

S(t’) = 0 = 80 + 15*t’ -0.5*9.81*(t’)^2

Solve this quadratic equation for t.

You’ll get two solutions, one of which is negative time; discard that one.

Fill in your solution for t’ into the velocity equation (2) and get the velocity as well.