Ball Dropping From Different Heights

Question

You drop a ball from height H. The speed of the ball right as it hits the ground (ie, the instant before it hits the ground) is V. You drop a ball from a height four times higher (4H). What is the velocity of the ball right as it hits the ground?

A) V

B) 2V

C) 4V

D) 8V

Answer 1

By energy conservation,

gH = v²/2 ⇒

v = √(2gH)

Let u be the final speed when drooped from 4H.

g(4H) = u²/2 ⇒

u = √[2g(4H)] = 2 √(2gH) = 2v

Answer 2

You drop a ball from height H. The speed of the ball right as it hits the ground (ie, the instant before it hits the ground) is V. You drop a ball from a height four times higher (4H). What is the velocity of the ball right as it hits the ground?

From height H,

H=(1/2)gt²

t=(2H/g)^1/2

V=-gt=-g(2H/g)^1/2

V=gg^-1/2(2H)^1/2

V=(2gH)^1/2

From height 4H,

4H=(1/2)gt²

8H=gt²

t=(8H/g)^1/2=2(2H/g)^1/2

U=gt=g[2(2H/g)^1/2]=2(2gh)^1/2=2V

2V