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
(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# , ground-level) -
#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 is
The ballon will move more 25 m
So Total height = (100+25)=125 m
Explanation:
Enjoy physics🙌🙌🙌:-) :-) :-)
