Question

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 21.8 m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 11.6 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 7.90 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 21.8 m/s .

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

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

Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.?

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

Apply kinematic equation  

H1 = V_0y*t - 1/2gt^2

= -21.8*7.9 - 1/2(9.8)(7.9)^2

= -478 m

While the stone was traveling H1 for 7.9s

H2 = V_0y*t = -7.9*21.8 = -172.22 m

H = H1-H2 = -478+172.22 = -305.78 m

The horizontal displacement is

X = Vxt = 11.6*7.9 = 91.64 m

The distance between the stone and the balloon is

d = sqrt(H^2 + x^2 )

= sqrt(305.78^2 + 91.64^2 )

= 319.21 m

The velocity components of the balloon

V_yr = V_by - V_py = (-gt)-0 = -9.8*7.9 = -77.42 m/s

Horizontal relative velocity is

Vxr = Vbx - Vpy = 11.8 - 0 = 11.8 m/s

Relative to person rest in the earth

Vyr = Vby - Vpy = -9.8*7.9 - 21.8 = -99.22 m/s

Vxr = Vbx -Vpy = 11.8 - 0 = 11.8 m/s