A 1.10 kg water balloon is shot straight up with an initial speed of 3.20 m/s. (

Question

A 1.10 kg water balloon is shot straight up with an initial speed of 3.20 m/s. (a) What is the kinetic energy of the balloon just as it is launched? (b) How much work does the gravitational force do on the balloon during the balloon's full ascent? (c) What is the change in the gravitational potential energy of the balloon–Earth system during the full ascent? (d) If the gravitational potential energy is taken to be zero at the launch point, what is its value when the balloon reaches its maximum height? (e) If, instead, the gravitational potential energy is taken to be zero at the maximum height, what is its value at the launch point? (f) What is the maximum height?

Explanation / Answer

For part a use the equation: KE=1/2mv^2 For part b use the equation: You need to find the distance using Vf^2=Vi^2+2ad where a=9.80m/s^, Vf=0 m/s, Vi=3.20 m/s, and solve for d Once, you find that, i believe Gravitational potential energy (PE)=mgh m=1.10 kg, g= 9.80m/s^2, and h= the distance you found in the previous equation. I hope that helps get you started!!

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