Oberlaudacu

## Answered question

2022-01-04

A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws stones vertically downward 1.0 s apart and observes that the cause a single splash. The initial speed of the first stone was 2.0 m/s.
a. How long after the release of the first stone does the second stone hit the water?
b. What was the initial speed of the second stone?
c. What is the speed of each stone as it hits the water?

### Answer & Explanation

boronganfh

Beginner2022-01-05Added 33 answers

Step 1
We will take downward is the positive direction of moving:
For the first stone, we can find its final velocity just before hitting the water.
The givens for the first stone:
${v}_{i}=2\frac{m}{s}$
${y}_{f}=50m$
${y}_{i}=0m$ because our origin is the upper point.
$a=9.8\frac{m}{{s}^{2}}$ Notice that the acceleration of gravity we are putting it positive because we considered the downward is positive. The velocity is increasing, So both g and v must have the same sign.
Then we will use ${y}_{f}={y}_{i}+{v}_{i}t+0.5a{t}^{2}$ to find the time.
$50=0+2t+0.5×9.8{t}^{2}$
$4.9{t}^{2}+2t-50=0$
$t=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$t=\frac{-2±\sqrt{{2}^{2}-\left(4×4.9×-50\right)}}{2×4.9}$
$t=\left(2.996,-3.4\right)s$ we reject the negative t
So ${t}_{\text{For the first stone}}=3.0s$
Step 2
The second stone had released after the first by one second and they reached the bottom at the same time.
So the time for the second stone is 2s.
${t}_{\text{For the second stone}}=2.0s$
Step 3
The initial speed for the second one:
${y}_{f}={y}_{i}+{v}_{i}t+0.5a{t}^{2}$
$50=0+\left({v}_{i}×2\right)+\left(0.5×9.8×{2}^{2}\right)$
${v}_{i}=\frac{50-19.6}{2}=15.2\frac{m}{s}$
${v}_{\text{i for the second stone}}=15.2\frac{m}{s}$
Step 4

${v}_{f1}=2+\left(9.8×3\right)=31.4\frac{m}{s}$
Step 5

${v}_{f2}=15.2+\left(9.8×2\right)=34.8\frac{m}{s}$

Neunassauk8

Beginner2022-01-06Added 30 answers

Step 1
The downward id the positive direction or moving
For the first stone
We can find the final velocity just before hitting the water
The given for the first stone
${v}_{i}=2\frac{m}{s}$
${y}_{f}=50m$
${y}_{i}=0m$
Thus our origin is the upper point
Step 2
a) The acceleration of gravity we are putting it positive because we considered the downward is positive the velocity is increasing
So both g and v must have the same sign
${y}_{f}={y}_{i}+{v}_{i}+0.5a{t}^{2}$
$50=0+2t+0.5×9.8{t}^{2}$
${4.9}^{2}+2t-50=0$
$t=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$=\frac{-2\sqrt{{2}^{2}-\left(4×4.9×-50\right)}}{2×4.9}$
$t=\left(2.996-3.4\right)s$
Thus t(for the first stone) $=3.0s$
The second stone had released after the first by one second and they reached the bottom at the same time
So the time for the second stone is 2s
Step 3
b) The initial speed of the second stone
${y}_{f}={y}_{i}+{v}_{i}+0.5a{t}^{2}$
$50=0+\left({v}_{i}×2t+0.5×9.8×{2}^{2}\right)$
${v}_{i}=\frac{50-19.6}{2}$
$=15.2\frac{m}{s}$
c) ${v}_{i}$ for the second stone is 15.2m/s
${v}_{f1}={v}_{i1}+a{t}_{1}$
$=2+\left(9.8×3\right)$
$=31.4\frac{m}{s}$
${v}_{f2}={v}_{i2}+a{t}_{2}$
$=15.2+\left(9.8×2\right)$
$=34.8\frac{m}{s}$

karton

Expert2022-01-11Added 613 answers

For the first stone:
The distance travelled is the initial velocity times time + distance for an acceleration from 0.
In this case, the acceleration is g (the acceleration of gravity $9.8m/{s}^{2}$)
So
$d={v}_{i}\cdot t+\frac{1}{2}\cdot g\cdot {t}^{2}$
Plug in the numbers:
$50=2\cdot t+\frac{9.8}{2}\cdot {t}^{2}$
Put in standard format:
$4.9{t}^{2}+2t-50=0$
not obviously factorable, so use the quadratic formula:
$t=\left(-2±\sqrt{\frac{{2}^{2}-4\cdot 4.9\cdot -50}{2\left(4.9\right)}}\right)$
$t=\frac{-2±\sqrt{984}}{9.8}$
$t=\frac{-2±31.4}{9.8}$
A negative value has no meaning so $t=\frac{29.4}{9.8}=3$ (This is the (a) answer)
This is the time after the 1st throw that BOTH stones hit the water, so the time for the second stone is 2 seconds.
Use the same equations to get the initial velocity of the 2nd stone. (This time, you'll be entering d and t and solving for vi )
For the velocity hitting the water, use $v={v}_{i}+g\cdot t={v}_{i}+9.8t$
plug in ${v}_{i}$ and t, and solve

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