Question

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?

Answer 1

`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

Answer 2

`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"`