If the initial height of a tennis ball in free fall is increased by a factor of 4, by what factor with the
final velocity increase at the point that it hits the ground?
(Ignore air redistance)
A) 2
B) 4
C)8
D) 1(i.e. the velocity will be the same)
E) The increase depends on the value of the initial height
Answer says A. May I know the explanation
Hi!
using s = 1/2(u+v)t
s * 4 = 4/2(u+v)t
s* 4 = (2u+2v)t
with v as final velocity
let me know if i made a mistake, hope this helps!
This makes sense!
I was trying to figure it out using GPE = KE, which also gave me v increase by factor 2:)
May I know your method as well
Lets assume:
mass = 1kg
gravitational acceleration = 10m/s^2
height = 1m
GPE = m * h * g = 1 * 1 * 10 = 10 J
if we increase height 4 times (4m):
GPE = 1 * 4 * 10 = 40 J
Just before the object falls from the height to ground, remember that all of its GPE turns into KE (so there is maximum velocity):
at original height 1m:
KE = (m * v^2) / 2 = 10 J
10 = (1 * v^2) / 2
v = √20 = 2√5 m/s
at 4 times height (4m):
KE = 40 J
40 = (1 * v^2) / 2
v = √80 = 4√5 m/s (which is 2x the velocity at original height:))