Question

Two students are on a balcony 20.7 m above the street. One student throws a ball (ball 1) vertically downward at 14.6 m/s; at the same instant, the other student throws a ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down. (a) What is the difference in the two ball's time in the air?
b) What is the velocity of each ball as it strikes the ground? c) what is the magnitude of ball one? is the direction upward or downward? d) what is the magnitude of ball 2? is the direction upward or downward? e) How far apart are the balls 0.800 s after they are thrown?

Answer 1

kinematic equtions:
1) Vf = Vi-g*t
2) Vf^2 = Vi^2+2*g*d
3) Yf = Yi + Vt-0.5*g*t^2
a)
Ball 1:
using 2) Vf = sqrt(14.6^2 + 2*9.8*20.7)=24.88
using 1) 24.88 = 14.6+9.8*t t=1.049 seconds
Ball 2:
using 3) 0 = 20.7 + 14.6*t - 4.9*t^2 t=4.028 second
difference = 4.028-1.049 = 2.98 seconds
B)
Ball 1 :
solved in part a. Vf = 24.88 m/s
Ball 2:
It will have the same speed. 24.88 m/s. This is due to the fact that after the ball is thrown upwards, it will reach it's highest point and begin to fall. When it reaches the height of the balcony again, it will be traveling at the same speed at which is was thrown upwards, being the same speed Ball 1 was thrown downwards at. So if you considered the starting point to be when it passes the balcony, both balls would have the same Vi. Can be shown the same way with kinematics.
C)
Assuming the question means as the balls hit the ground. The magnitude is the same as the speed, with a downward direction.
D)
Same goes for ball 2. as the final situation is the same.
E)
Ball 1: using 3)
Yf = 20.7 - 14.6(0.8)-0.5*g*(0.8)^2 = 5.88 ft above the ground
Ball 2: using 3)
Yf = 20.7 + 14.6(0.8)-0.5*g*(0.8)^2=29.24 ft above the ground.
Difference = 29.24-5.88 = 23.36 ft