A body is dropped from a height of 300 m exactly at the same instant another body is projected from the ground level vertically up with a velocity of 150 m per second find when they will meet
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Height from where the stone is dropped= 300m
= >g = 9.8 m/s2
=>velocity of projection of ground projectile = 150 m/s
=>Velocity of projection of the projectile from some height = 0
=>Let the particles meet at time t and at height h
=>Thus, height covered by the projectile from the ground
=>h= ut - ½ gt2
=>h= 150t -½ x 9.8 x t2
=>h = 150t – 4.9t2 … (1)
=>Distance covered by the projectile thrown down in time t will be, since it falls down from the rest u= 0
=>300 – h = ½ gt2
=>300 – (150t – 4.9t2 ) = 4.9t2 …by equation (1)
=>300= 150t
=>t=2s
=>Thus particles meet each other at 2s
Height at which particles meet=>
h => 150t – 4.9t2
h=> 150x2 – 4.9 x 4
=>h= 300-19.6
=> h= 280.4
= >g = 9.8 m/s2
=>velocity of projection of ground projectile = 150 m/s
=>Velocity of projection of the projectile from some height = 0
=>Let the particles meet at time t and at height h
=>Thus, height covered by the projectile from the ground
=>h= ut - ½ gt2
=>h= 150t -½ x 9.8 x t2
=>h = 150t – 4.9t2 … (1)
=>Distance covered by the projectile thrown down in time t will be, since it falls down from the rest u= 0
=>300 – h = ½ gt2
=>300 – (150t – 4.9t2 ) = 4.9t2 …by equation (1)
=>300= 150t
=>t=2s
=>Thus particles meet each other at 2s
Height at which particles meet=>
h => 150t – 4.9t2
h=> 150x2 – 4.9 x 4
=>h= 300-19.6
=> h= 280.4
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