Question

part a: A rock is thrown straight upwards from the edge of a bridge with an initial velocity of +35.0 m/s. What will be the velocity of the rock after 2.00 sec?part b: What is the displacement, Δ, at this time?

Answer 1

Part A. We are given that rock is thrown straight upwards from a bridge with an initial velocity of +35 m/s. To determine the velocity after 2 seconds we will use the following equation of motion for the velocity of a body in free fall:

`v_f=v_0-gt`

Where:

`\begin{gathered} v_f,v_0=\text{ final and initial velocities} \\ g=\text{ acceleration of gravity} \\ t=time \end{gathered}`

Now, we plug in the values:

`v_f=35(m)/(s)-(9.8(m)/(s^2))(2s)`

Now, we solve the operations:

`v_f=15.4(m)/(s)`

Therefore, the velocity after 2 seconds is 15.4 m/s.

Part B. To determine the displacement we will use the following formula:

`\Delta y=v_0t-(gt^2)/(2)`

Now, we substitute the values:

`\Delta y=(35(m)/(s))(2s)-((9.8(m)/(s^2))(2s)^2)/(2)`

Solving the operations:

`\Delta y=50.4m`

Therefore, the displacement is 50.4 meters.