The projectile launch relations allow to find the results for the maximum height of the jump of the cat is 0,, 287 m
The projectile launch studies the movement of the body near the earth, let's use the relationship
[tex]v_y^2 = v_o_y - 2 g y[/tex]
Where v_y is the vertical velocity, v_[oy} the initial vertical velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time
They indicate that the cat jumped with an initial velocity of v₀ = 2.74 m / s and an angle of θ = 60º, let's use trigonometry to find the components of the velocity
v₀ₓ = v₀ cos 60
[tex]v_o_y[/tex] = vâ‚€ sin 60
[tex]v_o_y[/tex] = 2.74 sin 60 = 2.37 m / s
In the point highest trajectory the vertical speed must be zero (v_y = 0), in the attached we can see a diagram of the speeds of the cat
0 = [tex]v_{oy}^2[/tex] - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
Let's calculate
y = [tex]\frac{2.37^2}{2 \ 9.8 }[/tex]
y = 0.287 m
In conclusion with the projectile launch relations we can find the results for the maximum height of the cat jump is 0,, 287 m
Learn more here: brainly.com/question/14290926
