Question

Jane is riding in a hot air balloon that is rising vertically at a constant speed of 3 m/s over a lake. She reaches out and drops a rock from the balloon when the distance from the rock to the water is 50 m. Use g = 10 m/s2, and let the up direction be positive. About how long after Jane drops the rock will it splash into the water?

Answer 1

The rock will splash into the water
`3.47s` after Jane drops it.

Step-by-step explanation:

Hi

• Known data

`v_(i)=3m/s, y_(i)=50, y_(f)=0m` and
`g=10m/s^(2)`.

• We are going to use the formula below

`y_(f)=y_(i)+v_(i)t-(1)/(2) gt_^(2)`

• Letting up direction be positive and computing with the known data, we have

`0=50+3t-(1)/(2) 10t_^(2)`

`0=50+3t-5t_^(2)` a second-grade polynomial, we obtain two roots, so
`t_(1)=3.47s` and
`t_(2)=-2.87s`, due negative root has no sense, we take the positive one, so the rock will splash into the water
`3.47s` after Jane drops it.