A balloon is flying up with a velocity of 5 m/s. at a height of 100 m a stone is dropped from it . at the instant the stone reaches the ground level the height of balloon will be ?
options :
1. 25 m
2. 0 m
3. 125 m
4. 100 m
please answer with full explanation
options :
1. 25 m
2. 0 m
3. 125 m
4. 100 m
please answer with full explanation
2 Answers
Answer:
(C)
Explanation:
Well, we're given that a stone is dropped from a balloon at height
What we can do first is find the time it takes the rock to reach the ground (i.e. height
#ul(y = y_0 + v_(0y)t  1/2g t^2#
where

#y# is the height at time#t# (#0# , groundlevel) 
#y_0# is the initial height (#100# #"m"# ) 
#v_(0y)# is the initial velocity (#5# #"m/s"# because it was traveling with the balloon) 
#t# is the time (what we're trying to find) 
#g# is the acceleration due to gravity (#9.81# #"m/s"^2# )
Let's solve the above equation for our unknown variable,
#t = (v_(0y) +sqrt((v_(0y))^2  4(1/2g)(y_0  y)))/(2(1/2g))#
(the
Plugging in known values:
#t = (5color(white)(l)"m/s" +sqrt((5color(white)(l)"m/s")^2  2(9.81color(white)(l)"m/s"^2)(100color(white)(l)"m"  0)))/(9.81color(white)(l)"m/s"^2)#
#= color(red)(ul(3.92color(white)(l)"s")#
#= color(red)(ul(4.94color(white)(l)"s")#
We take the positive solution, so
#color(red)(ulbar(stackrel(" ")(" "t = 4.94color(white)(l)"s"" "))#
Now what we do is find the position of the balloon at this time, using the equation
#ul(y = y_0 + v_(0y)t#
We have

#y = ?# 
#y_0 = 100# #"m"# 
#v_(0y) = 5# #"m/s"# 
#t = color(red)(4.94color(white)(l)"s"#
Plugging these in:
#y = 100color(white)(l)"m" + (5color(white)(l)"m/s")(color(red)(4.94color(white)(l)"s")) = color(blue)(ulbar(stackrel(" ")(" "124.94color(white)(l)"m"" "))#
Thus, it seems as if option (3) is correct.
Answer:
answer is
The ballon will move more 25 m
So Total height = (100+25)=125 m
Explanation:
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