If a ball is dropped on planet Krypton from a height of 20 " ft" hits the ground in 2" sec", at what velocity and how long will it take to hit the ground from the top of a 200 " ft"-tall building?

2 Answers
Jun 22, 2017

6.3246 seconds

Explanation:

Let accelaration due to gravity on Krypton be g_k.

Hence, when an object at a state of rest is dropped and it reaches ground in 2 seconds, then we have

S=1/2g_kt^2

or 20=1/2g_kxx2^2

or 20=2g_k and hence g_k=10 (ft)/(sec^2)

Let t be the time taken by the object to hit the ground from the top of a 200 ft tall building. Then

200=1/2xxg_kxxt^2

or 200=1/2xx10xxt^2

or 5t^2=200

or t=sqrt(200/5)=sqrt40=2sqrt10=2xx3.1623=6.3246 seconds

Jun 22, 2017

t~~6.32" s"
v~~63.2 "ft"/"s"

Explanation:

On Krypton, the velocity as a function of time is

v(t)=at

The position equation is the anti-derivative of the velocity, so

x(t)=int(at)dt

x(t)=1/2at^2+C

Let t=0 when the ball is first dropped, at height x(0)=20 feet.

1/2a(0)+C=20

C=20

So, the position equation becomes

x(t)=1/2at^2+20

When the ball hits the ground, the position will be x(2)=0 feet

1/2a(2)^2+20=0

2a=-20

a=-10 "ft"/"s"^2

You were asked for the time and velocity when the ball hits the ground from 200 feet.

x(t)=1/2(-10)t^2+200

1/2(-10)t^2+200=0

-5t^2=-200

t^2=sqrt(40)

t~~6.32 " sec"

The velocity when t=6.32 seconds is

v(6.32)=-10(6.32)=63.2 "ft"/"s"