Calculating Speed of Falling Banana - "Just Before" it Hits Ground?

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

"A monkey drops a banana. Its weight is 1N. It falls 3 metres to the ground. What is the speed of the falling banana just before it hits the ground?"

2. Relevant equations

Not sure, but I'm guessing I need to calculate the kinetic energy the banana gains (which was 3J) and the change in GPE (which was also 3J).

3. The attempt at a solution

My textbook says the answer is 7.8 m/s, but I have no idea why. As I've said above, I've worked out the change in GPE and kinetic energy the banana gains, but why is the answer 7.8 m/s and how are you supposed to arrive at this solution?

thebigstar25

before the banana hits the ground the velocity is max, which means that the kinetic energy is max so total energy at the moment is kinetic energy .. and at the point just before the monkey drops the banana the potential energy is max with value = mgh=1*3=3J so your total energy is 3J , thus the max. kinetic energy is 3J and you know that the kinetic energy = 0.5*m*v^2 (here m =0.1 kg) then substituting you will find that v =7.74 m/s

dayyou

thebigstar25 is correct, this is all about Conservation of Energy. Potential Energy at the start is equal to the Kinetic Energy at the bottom. And since you know the Force and Acceleration, you can figure out Mass. (F=ma) Then just plug it into the equation for PE (PE=mgh).