A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be the same height?

Question

51780 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T18:34:41

the time taken for the two balls to meet = $15m$ /25 m/s $= 3/5$ s

suppose the height is h and after a time t sec the balls meet.

the lower ball will travel a distance say h1 in time t secs

$h1= 25 .t - (1/2) g. t^2$

in the same time interval the upper ball will travel

15-h1 distance ,so

15 - h1 = $(1/2)g.t^2$

but you know h1 so substitute it

$15 - 25. t + (1/2) g t^2$ = $(1/2) g. t^2$

so finally you get 15 = 25. t

so time taken = $15 /25 sec$ = $3/5 sec.$

it is as if no g was working.