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 timet (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:
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