What is its maximum height and when it is attained? When will this ball reach the ground

Question

A ball is thrown upward from the top of a cliff 27 meters high at a velocity of 5.4 m/sec. Round to the nearest hundredth.

Answer 1

H = -4.9t²+vt + a

H = -4.9t² + 5.4t+ 27

H' = -9.8t + 5.4

0 = -9.8t + 5.4

9.8t = 5.4

t = 0.5510 (It will reach the height about a half second after being thrown)

H = -4.9*(0.551)²+(5.4)(0.551)+27

H = 28.48m (it is about 1.5 higher than where it was thrown)

0 = -4.9t² + 5.4t + 27

Solving the quadratic gives t=-1.96 (rejected) and t=2.96 seconds.

It will reach the height about a half second (.551 s) after being thrown, will reach an additional 1.48 m of height, and will start bouncing 2.96 seconds after being thrown

Answer 2

h(t) = (a/2)t^2 + vot + ho where a = acceleration, vo = initial velocity, t = time, ho = initial height

h = -4.9t^2 +5.4t + 27

take the derivative and set = 0

h' = -9.8t + 5.4 = 0, solve for t = time at max height

t =5.4/9.8 = 0.55 seconds plug back into h(t) equation to find max height

h(.055) = -4.9(5.4/9.8)^2 + 5.4(5.4/9.8) + 27 = -1.49 +2.98 + 27 = 28.49 meters = max height

It hits the ground when h(t) = 0

0 = -4.9t^2 + 5.4t + 27

use the quadratic formula to solve for t

t = -54/-9.8 + (1/9.8)sqr(5.4^2 +4(4.9)(27)) = 2.96 seconds to hit the ground