Question

A heavy balloon carrying a basket is moving with a constant downward velocity of 5.0 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 7.0m/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 4.0s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 5.0 m/s . How high was the balloon when the rock was thrown out? 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

Answer 1

Answer:98.4 m

Step-by-step explanation:

Given

Descending velocity of balloon(v)=5 m/s

velocity of stone relative to balloon(u)=7 m/s

thus u is the horizontal velocity of stone and v vertical velocity of balloon

time travel by stone to reach ground=4 s

`h=ut+(at^2)/(2)`

where u=initial velocity

here u=5 m/s

t=4 s

a=g

`h=5* 4+(9.8* 4^2)/(2)=98.4 m`

horizontal distance traveled in this time

`R=u* t`

`R=7* 4=28 m`

Thus stone at distance of 28 m from balloon

Also balloon and stone will be at same height at the time when rock hits ground