Question

a boll is thrown upward from the ground with an initial.speed of 25 m/s ,at the same instant , another boll is dropped from from a building 15 m.high . after how long will the bolls be at the same hight avobe the ground ?​

Answer 1

Correct question:-

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?

To find:-

• 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.

To find:-

• How long will the balls be the same height?

Answer:

The time taken for the two balls to meet = 15m/25m/s = $\tt\frac{3}{5}s$

Explanation:-

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.

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

in the same time interval the upper ball will travel

15-h1 distance.

Therefore,

$\tt15 - ht = ( \frac{1}{2} )g.t {}^{2}$

We know h1 so substitute it,

$\tt15 - 25.t + ( \frac{1}{2})gt {}^{2} = ( \frac{1}{2} )g.t {}^{2}$

so finally you get 15 = 25. t

So, time taken = $\tt\frac{15}{25}sec$ = $\tt\frac{3}{5}$

Therefore, Answer = $\tt\frac{3}{5}$