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An object is dropped from a height of 20m. What is its velocity just before hitting the ground? (Take g=10 m/s²)
20 m/s
10 m/s
40 m/s
25 m/s
20 m/s
Use the energy conservation shortcut: potential energy equals kinetic energy, so mg h =21 m v², which simplifies directly to v = 2gh.
Initial velocity u = 0 m/s, Height h = 20 m, Acceleration due to gravity g = 10 m/s²
v2=u2+2gh
Use the energy conservation shortcut: potential energy equals kinetic energy, so mg h =21 m v², which simplifies directly to v = 2gh.
Forgetting that the initial velocity u is zero for a dropped object or confusing distance with time.
Identify Given Values
Since the object is dropped, its initial velocity is u=0 m/s. The height is h=20 m and acceleration due to gravity is g=10 m/s2.
u=0 m/s,h=20 m,g=10 m/s2
Select Kinematic Equation
Use the third equation of motion which connects velocity, acceleration, and displacement without needing time.
v2=u2+2gh
Substitute and Calculate
Substitute the known values into the formula to find v2.
v2=02+2(10)(20)=400
Find Final Velocity
Take the square root of 400 to determine the final velocity.
v=400=20 m/s
A is correct because the calculated velocity just before hitting the ground is 20 m/s.
The relationship between gravitational potential energy and kinetic energy is fundamental for solving work-energy theorem problems in mechanics.
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