GED Science TestPhysical ScienceHard
A student drops a 0.5 kg ball from a height of 2 meters. Neglecting air resistance, what is the speed of the ball just before it hits the ground? (Assume g = 9.8 m/s²)
- A4.0 m/s
- B19.6 m/s
- C6.3 m/s
- D2.0 m/s
Show answer & explanationAnswer & explanation
Correct answer: C. 6.3 m/s
Using conservation of mechanical energy, potential energy (PE) at the top converts to kinetic energy (KE) at the bottom. PE = mgh and KE = 0.5mv². So, mgh = 0.5mv². The mass (m) cancels out, leaving gh = 0.5v². Solving for v: v = √(2gh) = √(2 * 9.8 m/s² * 2 m) = √(39.2) ≈ 6.26 m/s, which rounds to 6.3 m/s.
Why the other options are wrong
- A. Incorrect calculation; perhaps a misapplication of a formula.
- B. This is 2gh, not √(2gh).
- D. This would be the result if v = √(gh).
Conservation of Mechanical Energy (Free Fall)
In the absence of non-conservative forces like air resistance, the total mechanical energy (sum of potential and kinetic energy) of an object in free fall remains constant.
- PE_initial + KE_initial = PE_final + KE_final.
- For an object dropped, initial KE is zero, and final PE is zero (at ground).
- mgh = 0.5mv² can be used to find final velocity.
Memory trick: PE becomes KE as it falls, like a bank transferring funds.