A) Let's first find the time that the ball takes to reach the floor:
y = 1 2 g t 2 , y=\dfrac{1}{2}gt^2, y = 2 1 g t 2 , t = 2 y g = 2 ⋅ 2 m 9.8 m s 2 = 0.64 s . t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot2\ m}{9.8\ \dfrac{m}{s^2}}}=0.64\ s. t = g 2 y = 9.8 s 2 m 2 ⋅ 2 m = 0.64 s . Finally, we can find the distance from the edge of the table to the point where the ball lands:
x = v 0 t = 10 m s ⋅ 0.64 s = 6.4 m . x=v_{0}t=10\ \dfrac{m}{s}\cdot0.64\ s=6.4\ m. x = v 0 t = 10 s m ⋅ 0.64 s = 6.4 m . B) Let's first find final vertical velocity of the ball just before it strikes the floor:
v y = − g t = − 9.8 m s 2 ⋅ 0.64 s = − 6.27 m s . v_y=-gt=-9.8\ \dfrac{m}{s^2}\cdot0.64\ s=-6.27\ \dfrac{m}{s}. v y = − g t = − 9.8 s 2 m ⋅ 0.64 s = − 6.27 s m . Finally, we can find the ball’s velocity just before it strikes the floor from the Pythagorean theorem:
v = v x 2 + v y 2 = ( 10 m s ) 2 + ( − 6.27 m s ) 2 = 11.8 m s . v=\sqrt{v_x^2+v_y^2}=\sqrt{(10\ \dfrac{m}{s})^2+(-6.27\ \dfrac{m}{s})^2}=11.8\ \dfrac{m}{s}. v = v x 2 + v y 2 = ( 10 s m ) 2 + ( − 6.27 s m ) 2 = 11.8 s m .
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