# A body is thrown vertically upwards with a speed 29.4m/s.At the same time another body is dropped from a height 44.1m.When and where will they meet?

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### 1 Answer

Let us say the two objects meets at x meters from ground at time t.

For the object that thrown upward;

`uarr S = ut+1/2at^2`

`x = 29.4t-1/2*9.81*t^2` -----(1)

For the object that dropped downward;

`darr S = ut+1/2at^2`

`44-x = 0+1/2*9.81*t^2` -----(2)

(1)+(2)

`44 = 29.4t`

`t = 1.497S`

From (2);

`x = 33m`

*So the two objects will meet at 33m height from the ground and they will meet after 1.497 seconds after they were thrown/dropped.*

Assumption;

The upward throw starts at ground level

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