Answer:
[tex]h=60.025m[/tex]
Explanation:
Let's use the "Vertical displacement formula", which gives you the vertical displacement (Height) at any given time t:
[tex]\Delta y=y_f-y_i=v_o_yt-\frac{1}{2} gt^2[/tex]
Where:
[tex]y_f=Final\hspace{3}position=0\hspace{3}(because\hspace{3}has\hspace{3}reached\hspace{3}the\hspace{3}ground\hspace{3}height=0)\\y_i=Initial\hspace{3}position=h\hspace{3}(The\hspace{3}initial\hspace{3}position\hspace{3}represents\hspace{3}the\hspace{3}height\hspace{3}of\hspace{3}the\hspace{3}tower\\v_o_y=Initial\hspace{3}velocity=0\\t=time=3.5s\\g=Standard\hspace{3}gravity \approx-9.8m/s^2[/tex]
Hence:
[tex]-h=-\frac{1}{2} gt^2\\\\h=\frac{1}{2} gt^2[/tex]
Replacing the data provided by the problem:
[tex]h=\frac{1}{2} (9.8)*(3.5^2)=60.025m[/tex]
