A 124-kg balloon carrying a 22-kg basket is descending with a constant downward

Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of11.3 m/s . A 1.0-kg stone is thrown from the basket with an initial velocity of 13.1 m/sperpendicular 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 12.4 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 11.3 m/s .

Part A

How high was the balloon when the rock was thrown out?

Part B

How high is the balloon when the rock hits the ground?

Part C

At the instant the rock hits the ground, how far is it from the basket?

Part 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.

Part 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.

Explanation / Answer

this problem is related to the projectile motion.

Vb= 11.3 m/s

Us= 13.1 m/s

t=12.4 s

part A

from the second equation of motion

h=(Vb*t)+1/2*g*(t^2)

h=(11.3*12.4)+1/2*9.8*(12.4^2)

h=140.12+753.42

h=893.54 m

part B

h1= h- (Vb*t)

h1=893.54-140.12= 753.42 m

part C

D^2= (h1^2)+((Vs*t)^2)

D^2=((753.42)^2)+((13.1*12.4)^2)

D=sqrt(567569.36+26386.75)

D=770.68 m

part D

VX=13.1 m/s

Vy= (sqrt(2*g*h))-Vb

Vy=(sqrt(2*9.8*893.54 ))-11.3

Vy=132.33-11.3=121.03 m/s

part E

Vx=13.1 m/s

Vy=sqrt(2*g*h)

Vy=132.33 m/s

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