Question

If a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower?

Given: g = -9.8 meters/second2

Answer 1

34.3 meters
you multiply how many seconds bt 9.8m

Answer 2

`h=60.025m`

Step-by-step explanation:

Let's use the "Vertical displacement formula", which gives you the vertical displacement (Height) at any given time t:

`\Delta y=y_f-y_i=v_o_yt-(1)/(2) gt^2`

Where:

`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`

Hence:

`-h=-(1)/(2) gt^2\\\\h=(1)/(2) gt^2`

Replacing the data provided by the problem:

`h=(1)/(2) (9.8)*(3.5^2)=60.025m`