Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown from the basket with an initial velocity of 15.0 m/ s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume ,that the balloon continues its downward descent with the same constant speed of 20.0 m/ s. (a) How high was the balloon when the rock was thrown out? (b) How high is the balloon when the rock hits the ground') (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground

Answer 1

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$