Ignoring air resistance, a ball dropped from rest at altitude 250km strikes the ground with (a) 1500m/s (b) 2200m/s (c) 2800m/s

I already know that is 2200m/s but don't understand why???

From Vf^2 = Vo^2 + 2gh

Vf^2 = 0 + 2(9.8)250,000 making
Vf = 2213m/s

or

From h = Vo(t) = g(t^2)/2

250,000 = 9.8t^2/2 making t = 225.87sec.

and

Vf = Vo + gt

Vf = 0 + 9.8(225.87) = 2213m/s

To determine the final velocity of a ball dropped from an altitude, we can use the principle of conservation of energy. Ignoring air resistance, the only forces acting on the ball are gravity and the force exerted by the Earth.

The potential energy of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground. As the ball falls, its potential energy is converted into kinetic energy, given by the equation KE = (1/2)mv², where v is the final velocity of the ball.

Since energy is conserved, we can equate these two equations:
PE = KE

mgh = (1/2)mv²

By canceling out the mass m on both sides and simplifying the equation, we get:
gh = (1/2)v²

Solving for v, the final velocity of the ball after falling from altitude h is given by:
v = √(2gh)

In this case, the ball is dropped from a height of 250 km (or 250,000 m). The acceleration due to gravity, g, is approximately 9.8 m/s². Plugging in these values into the equation, we have:

v = √(2 * 9.8 * 250000)
v ≈ 2200 m/s

Therefore, the ball will strike the ground with a velocity of approximately 2200 m/s.