Question

You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time you drop your ball from the window. The two balls are initially separated by 28.7 m. (a) At what time do they pass each other? (b) At what location do they pass each other relative the window?

Answer 1

simple physics. the first point we call zero so it equ will be y=vf(t) and the other is already 28.7 down the line so its equ will be y=vf(t)+28.7........

Answer 2

well start with the fact that acceleration of gravity is about 9.8 m/s^2
then integrate to get velocity, then integrate again to get height ...(assuming you know calculus) its ok if you don't just use the formula
\[a(t) = -9.8\]
\[v(t) = -9.8t + v_{0}\]
\[h(t) = -4.9t^{2}+v_{0}t + h_0\]
to solve for v_f , you need to know velocity of first ball when h(t)=0
\[-4.9t^{2}+28.7 = 0\]
\[t = \sqrt{\frac{28.7}{4.9}}\]
plug this into velocity equation
\[v_f = 9.8 \sqrt{\frac{28.7}{4.9}}\]
**note, since v_f is speed , it doesn't matter if its negative
now v_f is initial velocity of 2nd ball...its height equation is
\[h(t) = -4.9t^{2}+(v_f)t\]
to find where the balls meet...set the 2 height equations equal
\[-4.9t^{2} +28.7 = -4.9t^{2} + 9.8\sqrt{\frac{28.7}{4.9}} t\]
solve for t
\[t = \frac{28.7}{9.8 \sqrt{\frac{28.7}{4.9}}}\]
this is the time when the 2 balls meet...plug this into either height equation to find the height
hope this helps