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

Question

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

2 Answers

Impact of this question

3166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-22T05:40:15

$6.3246$ $6.3246$ seconds

Explanation:

Let accelaration due to gravity on Krypton be $g_{k}$ $g_k$ .

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

$S = \frac{1}{2} g_{k} t^{2}$ $S=1/2g_kt^2$

or $20 = \frac{1}{2} g_{k} \times 2^{2}$ $20=1/2g_kxx2^2$

or $20 = 2 g_{k}$ $20=2g_k$ and hence $g_{k} = 10$ $g_k=10$ $\frac{f t}{{sec}^{2}}$ $(ft)/(sec^2)$

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

$200 = \frac{1}{2} \times g_{k} \times t^{2}$ $200=1/2xxg_kxxt^2$

or $200 = \frac{1}{2} \times 10 \times t^{2}$ $200=1/2xx10xxt^2$

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

or $t = \sqrt{\frac{200}{5}} = \sqrt{40} = 2 \sqrt{10} = 2 \times 3.1623 = 6.3246$ $t=sqrt(200/5)=sqrt40=2sqrt10=2xx3.1623=6.3246$ seconds

Answer 2 · 2017-06-22T17:17:58

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

Explanation:

On Krypton, the velocity as a function of time is

$v (t) = a t$ $v(t)=at$

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

$x (t) = \int (a t) d t$ $x(t)=int(at)dt$

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

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

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

$C = 20$ $C=20$

So, the position equation becomes

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

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

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

$2 a = - 20$ $2a=-20$

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

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

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

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

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

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

$t \approx 6.32 sec$ $t~~6.32 " sec"$

The velocity when $t = 6.32$ $t=6.32$ seconds is

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

Science

Math

Humanities

... and beyond

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

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If a ball is dropped on planet Krypton from a height of 20 " ft"20 ft20 " ft" hits the ground in 2" sec"2 sec2" sec", at what velocity and how long will it take to hit the ground from the top of a 200 " ft"200 ft200 " ft"-tall building?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

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