Question #23335

A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant a ball is dropped from rest from a building 14 m high. After how long will the balls be at the same height?

Expert's answer

Сoordinate of the first ball depends on time as follows:

x1 = 19.2t - 9.8t^2/2

19.2 - initial speed

9.8 - gravitational acceleration

Сoordinate of the second ball depends on time as follows:

14 - 9.8t^2/2

14 - initial coordinate

When the balls will be at the same height their coordinates will be equal:

x1 = x2:

19.2t - 9.8t^2/2 = 14 - 9.8t^2/2

t = 14/19.2 = 0.73 seconds

Answer: 0.73 sec

x1 = 19.2t - 9.8t^2/2

19.2 - initial speed

9.8 - gravitational acceleration

Сoordinate of the second ball depends on time as follows:

14 - 9.8t^2/2

14 - initial coordinate

When the balls will be at the same height their coordinates will be equal:

x1 = x2:

19.2t - 9.8t^2/2 = 14 - 9.8t^2/2

t = 14/19.2 = 0.73 seconds

Answer: 0.73 sec

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