A 3kg basketball is dropped from the top of a 120m building. if all of the energy is converted into KE, what is the velocity of the ball just before it hits the ground
potential energy = m g h = 3(9.81)(120)
Ke = .5 m v^2 = .5 (3)(v^2)
so
v^2 = 2 * 9.81 * 120
I am lost...please explain
well, yo9u can just memorize the formula. Most of us know it.
For something dropped with no air resistance from height h
v = sqrt (2 g h)
Let me see if I get this:
KE=1/2(m)(v^2)
but because I do not know the velocity, I would use the formula v=sqrt (2 g h)
v=sqrt (2 g h)
=sqrt (2*9.8*120)
=sqrt (2352)
then I would use the velocity 2352 for the formula
KE=1/2(m)(v^2)
=1/2 (3)(2352)
KE= 3528
You could do it the way I did it:
Potential energy lost = kinetic energy gained
m g h = (1/2) m v^2
g h = (1/2) v^2
2 g h = v^2
v = sqrt (2 g h)
However many students and teachers have just plain memorized
v = sqrt (2 g h)
because it is used so frequently.
Thank you so very much
To find the velocity of the ball just before it hits the ground, we can use the principle of conservation of energy. We know that the potential energy at the top of the building is converted into kinetic energy at the bottom. The potential energy (PE) of an object with mass (m) at a height (h) is given by PE = m * g * h, where g is the acceleration due to gravity (which is approximately 9.8 m/s²).
In this case, the potential energy at the top of the building is PE = 3 kg * 9.8 m/s² * 120 m = 3528 J (joules). This potential energy is converted completely into kinetic energy (KE) when the ball reaches the ground.
The kinetic energy of an object with mass (m) and velocity (v) is given by KE = 0.5 * m * v².
Now, we can equate the potential energy to the kinetic energy:
PE = KE
m * g * h = 0.5 * m * v²
Simplifying the equation, we can cancel out the mass (m) on both sides:
g * h = 0.5 * v²
Plugging in the known values, we have:
9.8 m/s² * 120 m = 0.5 * v²
Solving for v², we get:
v² = (9.8 m/s² * 120 m) / 0.5
v² = 2352 m²/s²
Taking the square root of both sides, we find:
v = √(2352 m²/s²)
v ≈ 48.49 m/s
Therefore, the velocity of the ball just before it hits the ground is approximately 48.49 m/s.