Two students are on a balcony 17.6 m above the street. One student throws a ball
ID: 1969633 • Letter: T
Question
Two students are on a balcony 17.6 m above the street. One student throws a ball vertically downward at 15.7 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. (Assume the positive direction is upward.)(a) What is the difference in the time the balls spend in the air?
______s
(b) What is the velocity of each ball as it strikes the ground?
ball thrown downward
______m/s
ball thrown upward
________m/s
(c) How far apart are the balls 0.800 s after they are thrown?
____m
Explanation / Answer
a)difference in the time that balls spend in air=time of flight of second body till it comes to same balcony position
=2u/g[since time taken by first ball=time taken by second after passing balcony both are equal]
=2*15.7/9.8=3.2041s
time taken by first ball to reach ground=(2h/g)=(2*17.6/9.8)=1.895s
b)velocity of each ball as it strikes the groung,V=U+at
=15.7+9.8*1.895
=34.27m/s[both will have same velocity on reaching ground]
c)t=0.8s
distance of first ball=ut+1/2at^2
=15.7*0.8+1/2*9.8*(0.8^2)
=15.696m[downward direction]
distance travelled by second body=-[ut+1/2at^2]
=-[15.7*0.8+1/2*(-9.8)*(0.8^2)]
=-9.424m(upwards)
distance apart by two bodies=15.696-(-9.24)
=25.12m
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