- #1

ironhawk316

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## Homework Statement

Question Details:

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=

speed=

distance=

## Homework Equations

y = yo + vot + 1/2 at2

Δy = Vot + 1/2 gt2

v2 = vo2 + 2g Δy

v = vo + gt

## The Attempt at a Solution

v = vo + gt

0=12 + (-9.8)t

-12= -9.8t

1.22=t

V2 = Vo2 +2g Δy

02 = (12)2 +2 (-9.8) Δy

02 = 144 + (-19.6) Δy

-144 = -19.6 Δy

7.35=Δy

Δy= Vot + 1/2 gt2

77.35 = o +1/2 (9.8)t2

77.35 = o +4.9 t2

15.79 = t2

3.97 = t

1.22 t + 3.97 t = 5.19 t

distance is 7.35m x 2 + 70m = 84.7 m

V= Vo = at

= -12 + 9.8 (5.2)

= -12 + 50.96

= 38.9 m/s

am I close or totally lost? Help please!