Two students are on a balcony 20.7 m above the street. One student throws a ball
ID: 1691750 • Letter: T
Question
Two students are on a balcony 20.7 m above the street. One student throws a ball vertically downward at 14.9 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. What is the difference in their time in air?
What is the velocity of each ball as it strikes the ground?
m/s
m/s
(c) How far apart are the balls 0.500 s after they are thrown?
m Two students are on a balcony 20.7 m above the street. One student throws a ball vertically downward at 14.9 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
. What is the difference in their time in air?
What is the velocity of each ball as it strikes the ground?
m/s
m/s
(c) How far apart are the balls 0.500 s after they are thrown?
m
Explanation / Answer
v0=v0y=v0*sin(90 degrees)=14.9 m/s y0=20.7 m t=v0y/g=(14.9 m/s)/9.8 m/s^2) approx = 1.52 s time for max height max height : y0 + voy^2/(2*g) = 14.9^2/(2*9.8)=11.327m + 20.7 m = 32.027 m from ground level ball thrown from balcony straight up when reaching max ht voy=0 m/s, ball will again attain voy=14.9 m/s level to balcony balcony to ground time : ( - 14.9 m/s + (14.9 ^2 + 4*1/2*9.8*20.7)^1/2)/( -2*4.9) = -1.03618 s total flight time for ball thrown up is 1.03618 + 1.52*2 = 4.07699 s approx = 4.08 s ball thrown from balcony straight dn time in air is 1.03618 s v on ground : (voy^2 original+voy^2 from balcony to grd^2)^1/2 c) distance apart at 0.5 s is : 1/2*9.8*0.5^2*2 = 2.45 m
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