A ball 'A' when thrown vertically upward reaches the balcony of a house 100m high. At the instant, when 'A' is thrown vertically upward, other ball 'B' is dropped from the balcony. When and where will the balls pass each other....??
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Answered by
0
d = ut+1/2gt^2
g= 10m/s^2
sqt20/2th
by which point the free falling ball will be fallen by 25m
Hence ans will be 75 m
g= 10m/s^2
sqt20/2th
by which point the free falling ball will be fallen by 25m
Hence ans will be 75 m
Answered by
1
Let the both meet at time t
A covers distance x in time t
and B covers distance 100-x in time t
let u be velocity of A with which it is projected
so when A is at 100 m its velocity is zero
so -u²=2(-10)(100)
u=10√20
so 100-x=1/2(10)t²=5t² ...(1) for B
and x=10√20t-5t²....(2)
add (1)+ (2)
100=10√20t
t=10/√20=√(10/2)=√5s
they meet after √5 secs
100-x=5(5)=25
x=100-25=75
so they meet at 75 m ground
A covers distance x in time t
and B covers distance 100-x in time t
let u be velocity of A with which it is projected
so when A is at 100 m its velocity is zero
so -u²=2(-10)(100)
u=10√20
so 100-x=1/2(10)t²=5t² ...(1) for B
and x=10√20t-5t²....(2)
add (1)+ (2)
100=10√20t
t=10/√20=√(10/2)=√5s
they meet after √5 secs
100-x=5(5)=25
x=100-25=75
so they meet at 75 m ground
Attachments:
onkie58:
@Manitkapoor2 ...... how did u get x=10√20t-5t² ...??..(2)
well from s=ut-1/2at^2
For A intial velocity u=10(20)^0.5 and a=g=10
so x=10(20)^0.5t-1/2(10)t^2
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