Two students are on a balcony 20.8 m above the street. One student throws a ball

Question

Two students are on a balcony 20.8 m above the street. One student throws a ball (ball 1) vertically downward at 16.6 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. What is the difference in the two ball's time in the air? s What is the velocity of each ball as it strikes the ground? ball 1 magnitude m/s direction ball 2 magnitude m/s direction How far apart are the balls 0.900 s after they are thrown? m When applying the equations of kinematics for an object moving in one dimension, which of the following statements must be true? The velocity of the object must remain constant. The position of the object must increase with time. The velocity of the object must always be in the same direction as its acceleration. The velocity of the object must increase with time.

Explanation / Answer

Ball 1 cover distance S1

S1 = ut + gt^2 / 2

S1 = 16.6*0.9 + 9.81*0.9^2 / 2

S1 = 18.913 m

Ball 2 cover distance S2

S2 = ut - gt^2 / 2

S2 = 16.6*0.9 - 9.81*0.9^2 / 2

S2 = 10.967 m

Balls apart = 18.913 - 10.967 = 7.946 m

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