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# A particle is dropped from tower of 2080 m.  At the same time second particle is thrown vertically in upward direction at 130 m/s. At what height they both will meet together. Please explain solution too.

Apoorva Arora IIT Roorkee
7 years ago
The time taken for both will be equal since they are thrown at the same instant. Also, The distance travelled by one will be 2080 minus the distance travelled by the other particle.
So we write the two eq of motion as
$1. s=0\times t+\frac{1}{2}\times 10\times t^{2}$
$2. 2080-s=130\times t-\frac{1}{2}\times 10\times t^{2}$
we get
2080=130t
hence t= 16 seconds
so height is 800 metres
Thanks and Regards
Apoorva Arora
IIT Roorkee