Respuesta :
Kinematics allows finding the height from where the stone was thrown 2.389m and the height measured from the door is Δy = 8.9 cm
given parameters
- time it takes for the stone to cross the door y = 0.28 s
- door height h = 2.3 m
to find
- height from where the stone was dropped (y₀)
The kinematics allows us to find the relationships between the position, velocity and acceleration of the stone. Let's fix the reference system at the bottom of the door.
y = y₁ + v₁ t - ½ g t²
where y and y₁ are the current and initial height of the stone, v₁ is the initial velocity, t is the time
They indicate that the stone is dropped and when it reaches the top of the door it has an initial velocity (v₁) and when it reaches the bottom of the setting its height is zero
0 = y₁ + v₁ t - ½ g t²
v₁ = [tex]\frac{\frac{1}{2} g t^2 - y_1}{t}[/tex]
we calculate
v₁ = [tex]\frac{\frac{1}{2} \ 9.8 \ 0.28^2 - 2.3 }{0.28}[/tex]
v₁ = -6.84 m / s
Now we can find out how high the stone is dropped so that it has this speed (v₁)
v₁² = v₀² - 2 g y₀
how the stone is released if initial velocity is zero
v₁² = 0 - 2g y₀
y₀ = [tex]\frac{v-1^2}{2g}[/tex]
y₀ = [tex]\frac{6.84^2}{2 \ 9.8}[/tex]
y₀ = 2.389 m
since the door has h = 2.3 m the height over this distance is
Δy = y₀ - h
Δy = 2.389 - 2.3
Δy = 0.089 m
The kinematics allows us to find the height from where the stone was thrown y₀ = 2.89 m and the height measured from the door is ΔY = 8.9 cm
learn more about kinematics here: brainly.com/question/10982199
