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!

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This makes sense!

I was trying to figure it out using GPE = KE, which also gave me v increase by factor 2:)

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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:))

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