Question

Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 16.8 m/s; at the same instant the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.

Answer 1

what is the question?

Answer 2

(a) What is the difference in their time in air?
s
(b) What is the velocity of each ball as it strikes the ground?
m/s
m/s
(c) How far apart are the balls 0.600 s after they are thrown?
m

Answer 3

The difference in their time will be the time it takes the ball of student 2 to go upwards up to its peak height and go downwards until it reaches the balcony level. Once the ball has reached the balcony level the time it takes to reach the ground is the same it takes for student´s 1 ball.
For student 2, the velocity of the ball is: \[v=v_0-g t\]and the peak height is reached when v=0\[v=0 \rightarrow t_{upwards}=v_0/g\]The time downwards (from peak height until balcony level) equals the time upwards therefore the total time is:\[t_{downwards}=t_{upwards} \rightarrow t_{difference}=2·\frac{ v_0 }{ g }\]b)Both balls will have the same velocity when hitting the ground. At balcony level they have a mechanical energy (potential+kinetic)\[E_M=mgh+\frac{ 1 }{ 2 }mv_0^2\]This energy at ground level (h=0), is only kinetic and has to be the same\[E_M=mgh+\frac{ 1 }{ 2 }mv_0^2=\frac{ 1 }{ 2 }mv_G^2\rightarrow v_G=\sqrt{v_0^2+2gh}\]
c) For Ball 1\[h_1=h_0-v_0t-\frac{ 1 }{ 2 }g t^2\]for Ball 2\[h_2=h_0+v_0t-\frac{ 1 }{ 2 }g t^2\]Distance will be:\[h_2-h_1=2v_0t\]