Answer:
y = 1.65 m
Explanation:
In this exercise they give the time to reach the assistant t = 1.15 s, let's use the kinematics equation
y = [tex]v_{oy}[/tex] t - ½ g t²
The initial vertical speed can be found with kinematics
sin θ = v_{oy} / v₀
v_{oy}= vo sin θ
Where tea is the launch angle of the egg
y = v₀ sin θ 1.15 - ½ 9.8 1.15²
y = v₀ sin θ 1.15 - 6.48
For a specific value you must give the initial velocity and the angle, if we assume that the velocity is v₀ = 10 m / s and the angle tea = 45
y = 10 sin 45 1.15 -6.48
y = 8.13 -6.48
y = 1.65 m
This height is approximately the part of a man's chest
