# 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?

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

$S = \frac{1}{2} {g}_{k} {t}^{2}$

or $20 = \frac{1}{2} {g}_{k} \times {2}^{2}$

or $20 = 2 {g}_{k}$ and hence ${g}_{k} = 10$ $\frac{f t}{{\sec}^{2}}$

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

$200 = \frac{1}{2} \times {g}_{k} \times {t}^{2}$

or $200 = \frac{1}{2} \times 10 \times {t}^{2}$

or $5 {t}^{2} = 200$

or $t = \sqrt{\frac{200}{5}} = \sqrt{40} = 2 \sqrt{10} = 2 \times 3.1623 = 6.3246$ seconds

Jun 22, 2017

$t \approx 6.32 \text{ s}$
$v \approx 63.2 \text{ft"/"s}$

#### Explanation:

On Krypton, the velocity as a function of time is

$v \left(t\right) = a t$

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

$x \left(t\right) = \int \left(a t\right) \mathrm{dt}$

$x \left(t\right) = \frac{1}{2} a {t}^{2} + C$

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

$\frac{1}{2} a \left(0\right) + C = 20$

$C = 20$

So, the position equation becomes

$x \left(t\right) = \frac{1}{2} a {t}^{2} + 20$

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

$\frac{1}{2} a {\left(2\right)}^{2} + 20 = 0$

$2 a = - 20$

$a = - 10 {\text{ft"/"s}}^{2}$

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

$x \left(t\right) = \frac{1}{2} \left(- 10\right) {t}^{2} + 200$

$\frac{1}{2} \left(- 10\right) {t}^{2} + 200 = 0$

$- 5 {t}^{2} = - 200$

${t}^{2} = \sqrt{40}$

$t \approx 6.32 \text{ sec}$

The velocity when $t = 6.32$ seconds is

$v \left(6.32\right) = - 10 \left(6.32\right) = 63.2 \text{ft"/"s}$