Suppose a baseball is thrown vertically upward from the ground with an initial velocity of Subscript[v, 0] ft/s. Its height above the ground after t seconds is given by s(t)=-16 t^2+Subscript[v, 0]t. Determine the initial velocity of the ball if it reaches a high point of 128 ft.

Respuesta :

Answer:

90.5

Step-by-step explanation:

Velocity is a time-derivative of displacement or the height, in this case.

[tex]s(t)=-16t^2+v_0t[/tex]

[tex]v(t) = -32t+v_0[/tex]

At maximum height, the velocity is 0 and the height is 128.

Substitute these values into both equations above,

[tex]0 = -32t+v_0[/tex]

[tex]t=\dfrac{v_0}{32}[/tex]

From the first equation,

[tex]128 = -16t^2 + v_0t[/tex]

Substitute for [tex]t[/tex].

[tex]128 = -16(\dfrac{v_0}{32})^2 + v_0\times\dfrac{v_0}{32}[/tex]

[tex]128 = -v_0^2/64 + v_0^2/32[/tex]

[tex]128=\dfrac{v_0^2}{64}[/tex]

[tex]v_0^2=8192[/tex]

[tex]v_0=64\sqrt{2}=90.5[/tex]

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