A particle is projected vertically upwards from a point O with initial speed 12.5 m s−1. At the same instant another particle is released from rest at a point 10 m vertically above O. Find the height above O at which the particles meet.?

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Answer 1 · 2018-04-09T08:22:35

The height is $=6.86m$

The acceleration due to gravity is $g=9.8ms^-2$

The height where the particles will meet is $=lm$

The distance travelled by the particle from the bottom is $=lm$

The distance travelled by particle from the top is $=(10-l)m$

Apply the equation of motion,

$s=ut+1/2at^2$

The time when they will meet is $=ts$

For the particle from the bottom

$l=12.5xxt-1/2xx9.8xxt^2$

For the particle from the top

$10-l=0xxt+1/2xx9.8xxt^2$

Therefore,

$10-(12.5xxt-1/2xx9.8xxt^2)=0xxt+1/2xx9.8xxt^2$

$10-12.5xxt+1/2xx9.8xxt^2=0xxt+1/2xx9.8xxt^2$

$10-12.5t=0$

$t=10/12.5=0.8s$

So,

$l=12.5*0.8-4.9*0.8^2=6.86m$