Answer:
The speed is v = 25 - 9.8 t
the velocity is v = (25 - 9.8 t) j^
the position is y = 25 t - 4.9 t²
Explanation:
This is a vertical throwing exercise.
The first thing we must do is set a coordinate system, in this case we will make the upward direction positive.
Let's write the kinematics equations
v = v₀ - g t
v² = v₀² - g (y-y₀)
y = y₀ + v₀ t - ½ g t²
in our case the initial velocity is
v₀ = 25 m / s
and we zero the system at the launch point
y₀ = 0
the equations remain
v = 25 - 9.8 t
v² = 25² - 9.8 y
y = 0 + 25 t - ½ 9.8 t²
The speed is v = 25 - 9.8 t
the velocity is v = (25 - 9.8 t) j^
the position is y = 25 t - 4.9 t²
