a stone is allowed to fall from top of the tower 50m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25m/s. calculate when the two stone will meet.
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A stone dropped from top will accelerate towards earth at 9.8 m/s^2
distance travelled by the stone is
s1=12at2s1=12at2
For the same stone, the distance travelled from ground is
s=50−s1=50−12at2s=50−s1=50−12at2
For the stone thrown upwards
s=25∗t−12at2s=25∗t−12at2
(minus sign because it is retardation)
When they meet, both distance from ground will be same. Therefore equating both we get
50−12at2=25∗t−12at250−12at2=25∗t−12at2
or
50=25∗t50=25∗t or t=2secondst=2seconds
distance from ground = 50−12∗9.8∗22=30.4m50−12∗9.8∗22=30.4m
distance from top = 50−30.4=19.6m50−30.4=19.6m
Therefore it takes 2 seconds for the stones to cross each other at a distance 19.6m from top of the tower.
distance travelled by the stone is
s1=12at2s1=12at2
For the same stone, the distance travelled from ground is
s=50−s1=50−12at2s=50−s1=50−12at2
For the stone thrown upwards
s=25∗t−12at2s=25∗t−12at2
(minus sign because it is retardation)
When they meet, both distance from ground will be same. Therefore equating both we get
50−12at2=25∗t−12at250−12at2=25∗t−12at2
or
50=25∗t50=25∗t or t=2secondst=2seconds
distance from ground = 50−12∗9.8∗22=30.4m50−12∗9.8∗22=30.4m
distance from top = 50−30.4=19.6m50−30.4=19.6m
Therefore it takes 2 seconds for the stones to cross each other at a distance 19.6m from top of the tower.
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