A stone is dropped from the upper observation deck of a tower, 100 m above the ground. (Assume g = 9.8 m/s2.)

Question

(a)

Find the distance (in meters) of the stone above ground at time t.

s(t) = 

 

(b)

How long does it take the stone to reach the ground? (Round your answer to two decimal places.)

_ s

(c)

With what velocity does it strike the ground? (Round your answer to one decimal place.)

 _m/s

(d)

If the stone is thrown downward with a velocity of 8 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

_s

Answer 1

s(t)=-4.9t^2 +100

=height of the stone from the ground at time t seconds

it hits the ground when s=0, when t^2=100/4.9 and t=10/sqr4.9 = about 4.52 seconds

v=s'(t)= -9.8t

v(10/sqr4.9) = -9.8(10/sqr4.9)= -98/sqr4.9= about 44.3 meters per second when it hits the ground

if thrown down at 8 m/s

then

s(t)= -4.9t^2 -8t +100

4.9t^2 +8t -100=0 where t= time when it hits the ground

use the quadratic formula, ignore the negative time

t= -8/9.8 +(1/9.8)sqr(64+4(490))

=-4/4.9+(sqr2024)/9.8

=about 3.77 seconds