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?

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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.