Sorin hits a cricket ball straight up into the air. After 1.5\,\text{s}1.5s1, point, 5, start text, s, end text, the ball is falling straight down with a speed of 6.5\,\dfrac{\text{m}}{\text{s}}6.5 s m 6, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We want to find the initial vertical velocity of the ball. We can ignore air resistance. Which kinematic formula would be most useful to solve for the target unknown? Choose 1 Choose 1 (Choice A) A v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t (Choice B) B {\Delta x}=\left(\dfrac{v+v_0}{2}\right)tΔx =( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t (Choice C) C \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared (Choice D) D v^2=v_0^2+2a\Delta xv 2 =v 0 2 +2aΔxv, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x (Choice E) E \Delta x=v t-\dfrac{1}{2}at^2Δx=vt− 2 1 at 2

What is the question? All I see is coding for a animation:
“Sorin hits a cricket ball straight up into the air. After 1.5\,\text{s}1.5s1, point, 5, start text, s, end text, the ball is falling straight down with a speed of 6.5\,\dfrac{\text{m}}{\text{s}}6.5 s m 6, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We want to find the initial vertical velocity of the ball. We can ignore air resistance. Which kinematic formula would be most useful to solve for the target unknown? Choose 1 Choose 1 (Choice A) A v=v_0+atv=v 0 +atv, equals, v, start subscript, 0, end subscript, plus, a, t (Choice B) B {\Delta x}=\left(\dfrac{v+v_0}{2}\right)tΔx=( 2 v+v 0 )tdelta, x, equals, left parenthesis, start fraction, v, plus, v, start subscript, 0, end subscript, divided by, 2, end fraction, right parenthesis, t (Choice C) C \Delta x=v_0 t+\dfrac{1}{2}at^2Δx=v 0 t+ 2 1 at 2 delta, x, equals, v, start subscript, 0, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared (Choice D) D v^2=v_0^2+2a\Delta xv 2 =v 0 2 +2aΔxv, squared, equals, v, start subscript, 0, end subscript, squared, plus, 2, a, delta, x (Choice E) E \Delta x=v t-\dfrac{1}{2}at^2Δx=vt− 2 1 at 2”

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