A stone is thrown Vertically

[Ironhawk]

9. Problem: A stone is thrown verticall upward with a speed of 12.0 m/s from the edge of a cliff 70.0m high. How much later does it reach the bottom of the cliff? What is its speed just before hitting the ground? What total distance did it travel?

time= 5.20s
speed= 38.9 m/s
distance=84.7m

I have a problem with calculating the time portion. I have found the answer but I am trying to make sence of how we arrived at the answer. This is the answer from the book but I don't understand the steps in getting 5.198


y=yo + vot + 1/2 at^2
(4.9m/s^2)t^2-(12.0m/s)t-70m=0
t=-2.749s, 5.198s

 

[Deleted member 4993]

9. Problem: A stone is thrown verticall upward with a speed of 12.0 m/s from the edge of a cliff 70.0m high. How much later does it reach the bottom of the cliff? What is its speed just before hitting the ground? What total distance did it travel?

time= 5.20s
speed= 38.9 m/s
distance=84.7m

I have a problem with calculating the time portion. I have found the answer but I am trying to make sence of how we arrived at the answer. This is the answer from the book but I don't understand the steps in getting 5.198


y=yo + vot + 1/2 at^2
(4.9m/s^2)t^2-(12.0m/s)t-70m=0 <<< This is a quadratic equation. The solution is probably obtained through quadratic formula.

The solution of quadratic equation:

A x[sup:3i7uy3gs]2[/sup:3i7uy3gs] + Bx + C = 0

x = [-B ± ?(B[sup:3i7uy3gs]2[/sup:3i7uy3gs] - 4AC)]/(2A)



t=-2.749s, 5.198s