Here we will use conservation of energy to tackle a simple free-fall problem. Yo
ID: 1451740 • Letter: H
Question
Here we will use conservation of energy to tackle a simple free-fall problem. You throw a 0.150 kgbaseball straight up in the air, giving it an initial upward velocity of magnitude 20.0 m/s. Use conservation of energy to find how high it goes, ignoring air resistance.
SET UP The only force doing work on the ball after it leaves your hand is its weight, and we can use the following equation:
12mvi2+mgyi=12mvf2+mgyf
We place the origin at the starting (initial) point (point i), where the ball leaves your hand; then yi=0 (Figure 1) . At this point, vi=20.0m/s. We want to find the height at the final point (point f), where the ball stops and begins to fall back to earth. At this point, vf=0 and yf is unknown.
SOLVE 12mvi2+mgyi=12mvf2+mgyf says that Ki+Ui=Kf+Uf, or
12mvi2+mgyi=12mvf2+mgyf
12(0.150kg)(20.0m/s)2+(0.150kg)(9.80m/s2)(0)
=12(0.150kg)(0)2+(0.150kg)(9.80m/s2)yf
yf=20.4m
The mass divides out, as we should expect; the motion of an object in free fall doesn’t depend on its mass.
REFLECT We could have substituted the values yi=0 and vf=0 into the following equation
12mvi2+mgyi=12mvf2+mgyf
and then solved algebraically for yf to get
yf=vi22g
(We could also have derived this result by using other methods, without energy considerations.)
QUESTION:
You throw the same baseball up from the surface of Mars (where the acceleration due to gravity is 3.71 m/s2) with the same speed as before. Find how high the ball goes
Explanation / Answer
on the Earth,
given, vo = 20 m/s
Apply conservation of energy
initial kinetic energy = fina potentail energy
(1/2)*m*vo^2 = m*g*h
==> h = vo^2/(2*g)
= 20^2/(2*9.8)
= 20.4 m <<<<<<<<-----------Answer
on the Mars,
given, vo = 20 m/s
Apply conservation of energy
initial kinetic energy = fina potentail energy
(1/2)*m*vo^2 = m*g*h
==> h = vo^2/(2*g)
= 20^2/(2*3.7)
= 54 m <<<<<<<<-----------Answer