A ball is projected vertically upward with a velocity of 30m/s. 2s later another ball is released from the top of a story building 80m tall. Determine after what time the balls bi pass each other and at what height do they bi pass?

1 Answer
Ananda Dasgupta
Mar 21, 2018

4 s after the second ball is dropped, at the ground (I have assumed #g=10" ms"^-2# for simplicity)

Explanation:

In 2s, the first ball will rise a distance of

#30" m/s"times 2" s"-1/2 times 10" ms"^-2 times 2^2 " s"^2=40" m"#

At this instant, its speed is

#30" m/s"-10" ms"^-2 times 2" s"=10" m/s"#

At this instant, the second ball is dropped. Now
- the relative separation of the two balls #=80" m"-40" m"=40" m"#
- their relative velocity #=10" m/s"#
- their relative acceleration #=0#

Thus the time it takes for the two to meet is #(40" m")/(10" m/s") = 4" s"#

In this time, the second ball will drop a distance of

#1/2 times 10" ms"^-2 times (4" s")^2=80" m"#

Thus the two balls will meet at the ground!

Related questions
Impact of this question
4446 views around the world
You can reuse this answer
Creative Commons License