Question

A stone is thrown vertically downward from a cliff 186.0-m tall. During the last half second of its flight, the stone travels a distance of 46.9 m. Find the initial speed of the stone. Answer must be in m/s

Answer 1

First,\[46.9 = u_2\cdot \dfrac{1}{2} + \dfrac{1}{2}\cdot 10 \cdot \left(\dfrac{1}{2}\right)^2\]\[u_2 = 2(46.9 - 1.25) = 2(45.65) = 91.3\]And then,\[u_2 ^2 =u_1^2 + 2\cdot 10 \cdot (186 - 46.9)\]

Answer 2

wait, so what's the difference between the two U's?

Answer 3

@ab60093

Answer 4

yes?

Answer 5

Assume the question as reverse. Suppose the ball is thrown upwards, now find u if t=1/2, s=46.9, a=-9.8

Answer 6

using what formula? Like, what's the base formula, I was never given one

Answer 7

Then using this u and \(\sf V^2=U^2-2aS\) find V, here take S=186 m. Value of V will be the answer.

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) @ab60093
using what formula? Like, what's the base formula, I was never given one
\(\color{blue}{\text{End of Quote}}\)

\(\sf S=ut + \frac{1}{2}at^2\)

Answer 9

U=96.25, so V=113.62?

Answer 10

recalculate v

Answer 11

i got the same answer. I typed it in as \[V=\sqrt{(96.25)^{2}-2(-9.8)(186)}\]

Answer 12

wait

Answer 13

i got 74.938

Answer 14

is that right?

Answer 15

finally !

Answer 16

:)

Answer 17

I guess u were using some kind of software before ?

Answer 18

no, I changed how I was typing it into my calculator. I think I was typing it in wrong

Answer 19

\(\color{blue}{\text{Originally Posted by}}\) @ab60093
i got the same answer. I typed it in as \[V=\sqrt{(96.25)^{2}-2(`-9.8`)(186)}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 20

you used negative sign two times !

Answer 21

lol. My bad. I did it as
v0 = √(96.255 m/s)2 + 2(–9.81 m/s2)(186.0 m)

Answer 22

then i got it right