Question

A hot-air balloon has just lifted off and is rising at the constant rate of 2.0m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?

Answer 1

Answer:3.085 m ,5.914 m

Step-by-step explanation:

Given

Velocity of hot air balloon(v)=2 m/s

Velocity at which camera is thrown up(u)=12 m/s

at t=0 passenger is at 2.5 m above

Let say it takes t sec for camera to reach balloon and balloon has traveled a distance of x m in that time

x=2t

also for camera

`2.5+x=12* t-(1)/(2)gt^2`------1

substitute value of x in 1

`2.5+2t=12t-5t^2`

`5t^2-10t+2.5=0`

`t^2-2t+0.5=0`

`2t^2-4t+1=0`

`t=(4\pm √(8))/(4)=(2\pm √(2))/(2)`

so there are two instances when both heights are equal

for
`t=(2-√(2))/(2)`

`x=2* (2-√(2))/(2)`

`x=2-√(2)=0.5857 m`

Therefore balloon is at a height of 2.5+0.5857=3.085 m

For
`t=(2+√(2))/(2)`

`x=2+√(2)`

balloon at a height of 5.914 m