Question

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.

Answer 1

Answer:

90.5

Step-by-step explanation:

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

$s(t)=-16t^2+v_0t$

$v(t) = -32t+v_0$

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

Substitute these values into both equations above,

$0 = -32t+v_0$

$t=\dfrac{v_0}{32}$

From the first equation,

$128 = -16t^2 + v_0t$

Substitute for $t$.

$128 = -16(\dfrac{v_0}{32})^2 + v_0\times\dfrac{v_0}{32}$

$128 = -v_0^2/64 + v_0^2/32$

$128=\dfrac{v_0^2}{64}$

$v_0^2=8192$

$v_0=64\sqrt{2}=90.5$