Oberlaudacu

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?

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?

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.

$50=0+2t+0.5\times 9.8{t}^{2}$

$4.9{t}^{2}+2t-50=0$

$t=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

$t=\frac{-2\pm \sqrt{{2}^{2}-(4\times 4.9\times -50)}}{2\times 4.9}$

$t=(2.996,-3.4)s$ 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:

$y}_{f}={y}_{i}+{v}_{i}t+0.5a{t}^{2$

$50=0+({v}_{i}\times 2)+(0.5\times 9.8\times {2}^{2})$

$v}_{i}=\frac{50-19.6}{2}=15.2\frac{m}{s$

$v}_{\text{i for the second stone}}=15.2\frac{m}{s$

Step 4

$v}_{f1}={v}_{i1}+a\text{}{t}_{1$

$v}_{f1}=2+(9.8\times 3)=31.4\frac{m}{s$

Step 5

$v}_{f2}={v}_{i2}+a\text{}{t}_{2$

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

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:

Then we will use

So

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.

Step 3

The initial speed for the second one:

Step 4

Step 5

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

$y}_{f}={y}_{i}+{v}_{i}+0.5a{t}^{2$

$50=0+2t+0.5\times 9.8{t}^{2}$

${4.9}^{2}+2t-50=0$

$t=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

$=\frac{-2\sqrt{{2}^{2}-(4\times 4.9\times -50)}}{2\times 4.9}$

$t=(2.996-3.4)s$

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

$y}_{f}={y}_{i}+{v}_{i}+0.5a{t}^{2$

$50=0+({v}_{i}\times 2t+0.5\times 9.8\times {2}^{2})$

$v}_{i}=\frac{50-19.6}{2$

$=15.2\frac{m}{s}$

c)$v}_{i$ for the second stone is 15.2m/s

$v}_{f1}={v}_{i1}+a{t}_{1$

$=2+(9.8\times 3)$

$=31.4\frac{m}{s}$

$v}_{f2}={v}_{i2}+a{t}_{2$

$=15.2+(9.8\times 2)$

$=34.8\frac{m}{s}$

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

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

Thus t(for the first stone)

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

c)

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

So

Plug in the numbers:

Put in standard format:

not obviously factorable, so use the quadratic formula:

A negative value has no meaning so

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

plug in

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