Answer:
It will take 1.25 s for the two balls to pass each other.
Explanation:
Given:
[tex]initial\ velocity_A\ (v_o_A)=0\ m/s[/tex]
[tex]initial\ height_A\ (h_o_A)=50\ m[/tex]
[tex]acceleration_A\ (a_A)=-9.8\ m/s^2[/tex] (acceleration going downward)
[tex]initial\ velocity_B\ (v_o_B)=40\ m/s[/tex]
[tex]initial\ height_B\ (h_o_B)=0\ m[/tex]
[tex]acceleration_B\ (a_B)=-9.8\ m/s^2[/tex] (acceleration going downward)
When the two balls passing each other, they will have the same time (t) and final height (h) → [tex]\boxed{h_A=h_B}[/tex]
[tex]\boxed{h=h_o+v_ot+\frac{1}{2}at^2 }[/tex]
[tex]h_A=h_B[/tex]
[tex]h_o_A+v_o_At+\frac{1}{2}a_At^2 =h_o_B+v_o_Bt+\frac{1}{2}a_Bt^2[/tex]
[tex]50+0t+\frac{1}{2} (-9.8)t^2=0+40t+\frac{1}{2} (-9.8)t^2[/tex]
[tex]50=40t[/tex]
[tex]t=\frac{50}{40}[/tex]
[tex]\bf t=1.25\ s[/tex]
