An astronaut on Jupiter drops a rock straight downward from a height of 1.05 m. If the acceleration of gravity on Jupiter is 24.8 m/s2, what is the speed of the rock just before it lands?
The Attempt at a Solution
At first i attempted to simply multiply 24.8 and 1.05, but realized that was probably wrong. I then calculated t by dividing 1.05 by 24.8 and got t=.042. Then used speed = distance/time to get s=1.05/.042 to get 1.04m/s. Does this sound correct, or is it way off? Would just like to know if this is the right method or if my answer is right. Thx