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 = $15 m$ $15m$ /25 m/s $= \frac{3}{5}$ $= 3/5$ 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

$h 1 = 25 . t - (\frac{1}{2}) g . t^{2}$ $h1= 25 .t - (1/2) g. t^2$

in the same time interval the upper ball will travel

15-h1 distance ,so

15 - h1 = $(\frac{1}{2}) g . t^{2}$ $(1/2)g.t^2$

but you know h1 so substitute it

$15 - 25 . t + (\frac{1}{2}) g t^{2}$ $15 - 25. t + (1/2) g t^2$ = $(\frac{1}{2}) g . t^{2}$ $(1/2) g. t^2$

so finally you get 15 = 25. t

so time taken = $\frac{15}{25} sec$ $15 /25 sec$ = $\frac{3}{5} sec .$ $3/5 sec.$

it is as if no g was working.