Question

Standing on a bridge, you throw a stone straight upward. The stone hits a stream, 32.5 m below the point atwhich you release it, 3.10 s later. What is the speed of the stone (in m/s) just after it leaves your hand? Pleasedo not include any units in your answer below. Type in only the numerical result. If you include units, youranswer will be marked as incorrect.

Answer 1

Given:

Distance the stone hits the stream = 32.5m below the released point

Time = 3.10 seconds

Let's find the speed of the stone just after it leaves your hand.

To find the speed of the stone, apply the kinematic formula:

`\Delta y=v_(iy)\ast t-(1)/(2)g\ast t^2`

Since the point the stone hits the stream is below the released point is, the change in distance is:

`\Delta y=0-32.5=-32.5m`

Where:

a = -g = -9.8 m/s^2

t = 3.10 s

Substituet values into the formula and solve for the speed of the stone (vy).

We have:

`\begin{gathered} -32.5=v_(iy)\ast3.10-(1)/(2)(9.8)\ast3.10^2 \\ \\ -32.5=v_(iy)\ast3.10-4.9\ast9.61 \\ \\ -32.5=v_(iy)\ast3.10-47.089 \end{gathered}`

Solving further:

`\begin{gathered} v_(iy)\ast3.10=-32.5+47.089 \\ \\ v_(iy)\ast3.10=14.589 \end{gathered}`

Divide both sides by 3.10:

`\begin{gathered} (v_(iy)\ast3.10)/(3.10)=(14.589)/(3.10) \\ \\ v_(iy)=4.706\text{ m/s} \end{gathered}`

Therefore, the speed of the stone just after it leaves your hand is 4.706 m/s

ANSWER:

4.706 m/s