CALCULUS HW PLEASE HELP LAST QUESTION

Question

A stone is dropped from the upper observation deck of a tower, 300 m above the ground. (Assume g = 9.8 m/s².)

1. How long (in seconds) does it take the stone to reach the ground? (Round your answer to two decimal places.)
2. With what velocity (in m/s) does it strike the ground? (Round your answer to one decimal place.)
3. If the stone is thrown downward with a velocity of 4 m/s, how long (in seconds) does it take to reach the ground? (Round your answer to two decimal places.)

Answer 1

h(t) = -4.9t^2 + 300 is the height or distance function, with zero initial velocity and initial height of 300 meters.

it hits the ground when h(t) = 0 = -4.9t^2 +300

t^2 = 300/4.9 = 61.22

t =sqr(61.22) = 7.8246 = 7.82 seconds

velocity is the derivative of the distance function or h'(t) = -9.8t

h'(7.8246) = -9.8(7.8246) = 76.7 meters per second= velocity as it hits the ground

It went from zero to 76.7 meters per second, which averages to almost 40 m/sec. 300 meters at almost 40 meters per second is slightly more than 30/4 = 7.5 seconds

If initial velocity had been -4 meters per second the distance function would have been

h(t) = -4.9t^2 -4t + 300

set = 0 and solve for t to find when it hits the ground

4.9t^2 + 4t -300 = 0

use the quadratic formula, ignore the negative square root

t = -8/9.8 + (1/9.8)sqr(16+4(4.9)(300) = -.816 + 7.835 = 7.02 seconds