(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A passenger is rising in a hot air balloon at a constant 2m/s. The passenger realizes she forgot her camera. A second person on the ground below the balloon throws the camera straight up when the balloon is 2.5m off the ground. What is the minimum velocity the camera can be thrown with in order to just reach the passenger?

2. Relevant equations

[tex]y_1(t)=2t+2.5[/tex]

[tex]y_2(t)=vt-\frac{1}{2}gt^2[/tex]

3. The attempt at a solution

Okay so I figured that at the minimum speed the camera will reach the passenger when its velocity is zero.

So figured out that its velocity is zero when [tex]t=\frac{v}{g}[/tex]

okay then I set the two position equations equal getting [tex]2t+2.5=vt-\frac{1}{2}gt^2[/tex]

So then my next step I wasn't sure was valid, but I substituted [tex]\frac{v}{g}[/tex] for t and then solved for v.

I got 9.28 m/s.

I don't know if this answer is correct or not.

Can someone please confirm my methods?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Minimum speed kinematics

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