Question

A hot air ballon is projected vertically with a net vertical acceleration of 4 m/s2

, by lighting

its burner. The burner is blown out in 2 minutes but the ballon continues to move up. Find

the maximum height reached by the balloon before it starts to come down again​

Answer 1

Answer:

Explanation:

Given, net vertical acceleration, a`=4m//s^(2)`

`h_(1)=ut+(1)/(2)at^(2)`

`=0+(1)/(2)xx4xx(120)^(2)`

=28,800 m

velocity attained by the hot air balloon after 2 min (120s).

v=u+at

`=0+4xx120`

=480m/s

After the burner is blown out,

u=480m/s and final velocity, v=0

Using `v^(2)-u^(2)=2gh_(2)`, we get

`rArr h_(2)=((480)^(2))/(2xx9.8)=11,755m`

`therefore` Maximum height reached by the balloon is

`h=h_(1)+h_(2)=(28,800+11,755)m`

`=40,555m=40.555`. km

=41km