Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 24.3 m/s . A 1.0-kg stone is thrown from the basket with an initial velocity of 12.9 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 11.8 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 24.3 m/s .
Required:
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 at rest in the basket.
e. Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest on the ground.

Answer 1

Answer:

-969.06

-286.74

698.7

-115.6, 12.9

-139.9, 12.9

Explanation:

Given that

Speed v, wrt y = -24.3 m/s

Speed v, wrt x = 12.9 m/s

time t, = 11.8 s

a

Using the formula

H(t) = ut - 1/2gt², where u = v wrt y

H(t) = -24.3 * 11.8 - 1/2 * 9.8 * 11.8²

H(t) = -286.74 - 682.28

H(t) = -969.06 m

b

H = ut, where u = v wrt y

H = -24.3 * 11.8

H = -286.74 m

H(1) = -969.06 - -286.74 = -682 m

c

Horizontal displacement, x = vt. Where v = v wrt x

x = 12.9 * 11.8

x = 152.22 m

d = √(H1² + x²)

d = √682² + 152²

d = 465124 + 23104

d = √488228

d = 698.7 m

d

Vertical component =

-gt - 0 =

-9.8 * 11.8 = -115.6

Horizontal component =

v wrt x - 0

12.9 - 0 = 12.9

e

Vertical component =

-gt - v wrt y =

-9.8 * 11.8 - 24.3 = -139.9

Horizontal component =

v wrt x - 0 =

12.9 - 0 = 0