Two students are on a balcony 20.0 m above the street. One student throws a ball
ID: 1571477 • Letter: T
Question
Two students are on a balcony 20.0 m above the street. One student throws a ball (ball 1) vertically downward at 15.8 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.How far apart are the balls 0.700 s after they are thrown?
ball 1 magnitude m/s direction -- ball 2 magnitude m/s direction --Explanation / Answer
a) for the ball thrown downward,
let's find the velocity with which it hits the ground
v^ 2= 15.8^2 + 2 (9.8)( 20)=
v = 25.33 m/s
time= 0.972 sec apprx
b) for the ball thrown upward,
max heigh reached = 15.8^2/ ( 19.6) = 12.7 m apprx
time to reach the max height= 15.8/9.8= 1.61 sec apprx
time to cover an height of 20 + 12.7 = sqroot ( 2h/g) = 2.15 sec
a) difference in timings= ( 1.61 + 2.15 - 0.972) = 2.79 sec apprx
b) velocity with which they hit the fllor= 25.33 and sqroot ( 2x9.8 x 32.7) = 25. 316 m/s
c) distance covered by downward ball after 0.7 sec = h = 15.8 ( 0.7) + 0.5 (9.8)(0.7^20 = 13.461 m
disatnce covered by upward ball in o,7 sec= 8.659 m
diffeence= 22.12 m apprx
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