The kinematics allows finding the result for the height of the cliff so that the bodies go down together is: 1,533 m
Kinematics studies the movement of bodies, finding relationships between the position, velocity and acceleration of bodies
In this case the bodies are thrown vertically so their acceleration is the gravity acceleration, directed downwards.
Let's set a reference system with the positive upward direction and zero at the base of the cliff, let's use the subscript 1 for the body that comes out from the bottom of the cliff and the subscript 2 for the body that comes out of the top, the equations for each body they are:
body 1
[tex]v_1^2 = V_{o1}^2 - 2g (y_1-y_{o1} )[/tex]
body 2
[tex]v_2^2 = v_{o2}^2 - 2g (y_2-y_{o2})[/tex]
In this case they indicate the initial velocities of each body, [tex]v_{o1}[/tex] = 80 m/s and [tex]v_{o2}[/tex] = 50 m/s, the initial height [tex]y_{o1}[/tex] = 0 and [tex]y_{o2}[/tex] = h, when the bodies fall at the same time they must begin to descend from the same height
y₁ = y₂
The velocities at this point must be zero
v₁ = v₂ = 0
We substitute
0 = [tex]v_{o1}^2 - 2 g y[/tex]
0 = [tex]v_{o2}^2 - 2 g ( y + h)[/tex]
Let's solve the system of equations
[tex]h = \frac{v_{o1} - v_{o2}}{2g}[/tex]
Let's calculate
h = [tex]\frac{80-50}{2 \ 9.8}[/tex]
h = 1.53 m
Consequently, the kinematics allows finding the result for the height of the cliff so that the bodies go down together is: 1.53 m
Learn more here: brainly.com/question/17772577
