Two students are on a balcony 20.7 m above the street. One student throws a ball
ID: 2282262 • Letter: T
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?
Explanation / Answer
a) the difference in time is equal to the time taken by the ball thrown vertically upwards to reach the balcony
we use
y = vt + 0.5 a t2
when the ball reaches balcony , the displacement y =0
v = 14.7m/s
a = -9.8m/s2
0 = 14.7 t - 4.9t2
t = 1.73 s
b) both ball will hit the ground with same velocity
we consider the ball thrown downwards
vf2 = vi2 + 2ay
vi = -14.7 m/s
a = -9.8m/s2
y = -20.7m
Vf = 24.94m/s
c) ball one would strike graound at 24.94 m/s downward
d) ball two would strike ground at 24.94m/s downward
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