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#