1 Apply the equation for the position as a function of time, but this time in the y-direction (vertical direction). Determine and plug in the y-component of the initial velocity and acceleration in the y-direction to find the relative height at the instant the egg reaches the horizontal position of the assistant (t1.15 s). So does the egg hit the assistant? If it does, at what part of his body?

   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 = 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

this could be newton#'s law of cooling. the cooling curve is exponential