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A hot-air balloon is rising straight up with a speed of 3.0 m/s. A bal-last bag is released from rest relative to the balloon at 9.5 m above the ground. How much time elapses before the ballast bag hits the ground?

$t=1.732 \mathrm{s}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

10:12

A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.

02:00

A hot-air balloon is risin…

00:40

02:29

03:22

During liftoff, a hot-air …

01:55

A hot-air balloon rises fr…

01:39

A balloonist, riding in th…

02:39

A hot-air balloon is ascen…

05:12

02:38

Free-Fall Acceleration…

So the question states that a hot air balloon is rising at three meters per second and at a height of 9.5 meters it drops a bag of some sort. And we're trying to find how long it takes for the bag to hit the ground. So to do this we can use our kingdom attic equations, one of which states that the change in displacement is vehicle to the initial blast e times the time plus 1/2 times the acceleration times the time squared. We know our displacement is gonna be negative 9.5 meters. We know that the initial velocity of the bag is gonna be three meters times. The time we're saying down in this case is negative. Of course, that's why this policeman is negative and of lost is positive. Then we add 1/2 times the acceleration which is negative 9.8 due to gravity multiplied by Times Square. And when we simplify this, we see that this equation is same thing as negative 4.9 times T squared plus three t plus 9.5 is equal to zero. And we can either use quote the quadratic formula solve for T, or we could just plug it into a calculator, which I did here, and we see that our route that's important. We don't want to look at the negative route because negative time doesn't make sense. But our positive route is 1.732 and that is our time t, and that's the answer to the problem.

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