body is dropped from a height of 100m Cliff and at the same time another body is thrown from the ground with 25 m per second velocity in upward direction where the Two will meet( take Gas 10 M per second)
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Can you please solve one more numrical
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hey dear your answer is here..
part 1
h=100m
u=0
let objects drops x meter
and time is t
part 2
u=25 m/s
distance upwards is 100-x
time is t
now
for part 1
s=ut +(at^2)/2
x=0×t+(10 t^2)/2
x=5t^2...(1)
part 2
s=ut +(at^2)/2
100-x=25t+(10 t^2)/2
100-x=25t+5t^2.....(2)
from 1 put value of x in 2
100-5t^2=25t+5t^2
10t^2+25t-100=0
2t^2+5t-20=0
part 1
h=100m
u=0
let objects drops x meter
and time is t
part 2
u=25 m/s
distance upwards is 100-x
time is t
now
for part 1
s=ut +(at^2)/2
x=0×t+(10 t^2)/2
x=5t^2...(1)
part 2
s=ut +(at^2)/2
100-x=25t+(10 t^2)/2
100-x=25t+5t^2.....(2)
from 1 put value of x in 2
100-5t^2=25t+5t^2
10t^2+25t-100=0
2t^2+5t-20=0
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