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

Answer:

#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:

#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"#