What is this problem's maximum height?

Question

Can you show or explain to me how you do this.

This is a three part question. But i need the maximum height to the three problems right?

A stone is thrown vertically upward with a speed of 13.0 m/s from the edge of a cliff 78.0 m high

I got 63.7 m
But i think i have to add it to the height of the cliff= 78 m +63.7 m= 141.7m (maximum height?)

(a) How much later does it reach the bottom of the cliff?
s
(b) What is its speed just before hitting?
m/s

Answer 1

Okay, i got something else.

This is just assuming it's going straight vertically:

Vi = 13
Vf=0
a= -9.8
di= 78
df= ?
t= ?

Vf is 0 because it's at maximum height, with no velocity at that instant T.
Now we have to find t, so we can use Vf= Vi + at.
0 = 13 + (-9.8)t
-13/-9.8 = 1.326 = T

then we can use that Displacement formula: Df - Di = 1/2(a)(T^2) + Vi(T)

plug in:
Df - 78 = -4.9(1.326^2) + 13(1.326)
Df - 78 = 8.622
df = 86.622

That should be your maximum height from the bottom of the cliff.

Answer 2

Using v^2 = u^2 + 2as
where v=13, u=0, a= -9.81
s=8.61

total height is 78+8.61=86.61

Answer 3

v^2 = vo^2 + 2adY

dY = (v^2 - vo^2) / 2a

dY = (0 - 169) / 2(-9.80)

dY = 8.62 m

8.62 m is the how high the stone went when you threw it. The max. height is 8.62 + 78, which equals 86.6 m.