A particle is dropped from a height of
100 m and another particle is projected vertically up with velocity 50 m/s from
the ground along the same line. Find out the position where two particles will
meet?
Let the upward direction is positive. And imagine the particles will meet at a
distance y from the ground.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiiF2ha5m4Tm84tuG0R38wkmA0SqDnP6Ns76lobB7p67L9SNJbFs6h-OnYQMkHRkIGfhak9ut8Asd5K2Y8mgM6wYMzIxyzgfZKjx9wKAtXJU5HwgRXNIvCDsKH4cz0zSkV_9drGkGXndMs/s1600/eqn+EQN1.png)
For particle A,
Y0
= +100 m
u = 0 m/sec
a = -10 m/sec2
We
know,
Y0
= 0 m
u = +50 m/sec
a = -10 m/sec2
Also,
From (1) and (2), by
equating, we will get
t = 2 sec.
By putting t =2 sec in
equation (1) or (2), you will get, y =80 m.
It means, the particle
will meet at a height of 80 m from ground.
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