The initial height from which the stone was dropped is required.
The initial height from which the stone was dropped was 239 feet above the ground.
Kinematic equations of linear motion
[tex]s[/tex] = Displacement = [tex]230-14=216\ \text{feet}[/tex]
[tex]u[/tex] = Initial velocity
[tex]t[/tex] = Time taken = 3 s
[tex]g=a[/tex] = Acceleration due to gravity = [tex]32\ \text{ft/s}^2[/tex]
From the kinematic equations we have
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow u=\dfrac{s-\dfrac{1}{2}at^2}{t}\\\Rightarrow u=\dfrac{216-\dfrac{1}{2}\times 32\times 3^2}{3}\\\Rightarrow u=24\ \text{ft/s}[/tex]
Now [tex]u=0,v=24\ \text{ft/s}[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{24-0}{32}\\\Rightarrow t=0.75\ \text{s}[/tex]
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}\times 32\times 0.75^2\\\Rightarrow s=9\ \text{ft}[/tex]
Initial height from where the stone was dropped is [tex]230+9=239\ \text{ft}[/tex]
Learn more about kinematic equations of motion:
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