Two students are on a balcony 20.1 m above the street. One student throws a ball
ID: 1957926 • Letter: T
Question
Two students are on a balcony 20.1 m above the street. One student throws a ball (ball 1) vertically downward at 15.9 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?
______ s
b. What is the velocity of each ball as it strikes the ground?
ball 1 magnitude ______ m/s
direction ______ (east, west, south, or north)
ball 2 magnitude ______ m/s
direction ______ (east, west, south, or north)
c. How far apart are the balls 0.700 s after they are thrown?
_______ m
Explanation / Answer
a) Ball 1 hits the ground in a time of t1 = 20.1/159 = 1.2641 sec
Ball 2 reaches the peak in a time of t = u/g = 15.9/9.81 = 1.6208 sec
Ball 2 reaches above the balcony a height of h = ut - (1/2)gt2 = 15.9*1.6208 - (1/2)*9.81*1.62082 = 12.8853m
Total height = 20.1+12.8853 = 32.9853m
Total time taken by ball 2 to reach ground = 1.6208 + (2*32.9853/9.81) = 4.214 sec
Difference in two ball's time in air = 4.214 - 1.2641 = 2.95 sec
b) We know that v2-u2 = 2as --->
Ball 1 hits the ground at a velocity of v = (15.92 + 2*9.81*20.1) = 25.44 m/s towards south
Ball 2 hits the groung at a velocity of v = (2*9.81*32.9853) = 25.44 m/s towards south
It means both balls hit with same velocity
c) Location of ball 1 after 0.7 sec from ground = 20.1 - (15.9*0.7 + (1/2)* 9.81*0.72) = 6.566 m
Location of ball2 after 0.7 sec from ground = 20.1 + (15.9*0.7 - (1/2)* 9.81*0.72) = 28.82655 m
After 0.7 sec they are at a distance of 22.26 m
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