Answered

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 :

lestat345

Answer:

The rock will splash into the water [tex]3.47s[/tex] after Jane drops it.

Explanation:

Hi

  • Known data

[tex]v_{i}=3m/s, y_{i}=50, y_{f}=0m[/tex] and [tex]g=10m/s^{2}[/tex].

  • We are going to use the formula below

[tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2} gt_^{2}[/tex]

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

[tex]0=50+3t-\frac{1}{2} 10t_^{2}[/tex]

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

Other Questions