Answer:
5010 m
Explanation:
The vertical position of the teacher at time t can be found by using the equation:
[tex]y(t) = h + ut+ \frac{1}{2}gt^2[/tex]
where
h is the initial height
[tex]u = -20 m/s[/tex] is the initial velocity (negative since it's downward)
[tex]g=-9.8 m/s^2[/tex] is the acceleration of gravity
t is the time
The teacher reaches the ground when y = 0, so the equation becomes
[tex]h=-ut-\frac{1}{2}gt^2[/tex]
Substituting t = 30 s, we find the initial height:
[tex]h=-(-20)(30)-\frac{1}{2}(-9.8)(30)^2=5010 m[/tex]
