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:
[tex]v_f=v_0-gt[/tex]
Where:
[tex]\begin{gathered} v_f,v_0=\text{ final and initial velocities} \\ g=\text{ acceleration of gravity} \\ t=time \end{gathered}[/tex]
Now, we plug in the values:
[tex]v_f=35\frac{m}{s}-(9.8\frac{m}{s^2})(2s)[/tex]
Now, we solve the operations:
[tex]v_f=15.4\frac{m}{s}[/tex]
Therefore, the velocity after 2 seconds is 15.4 m/s.
Part B. To determine the displacement we will use the following formula:
[tex]\Delta y=v_0t-\frac{gt^2}{2}[/tex]
Now, we substitute the values:
[tex]\Delta y=(35\frac{m}{s})(2s)-\frac{(9.8\frac{m}{s^2})(2s)^2}{2}[/tex]
Solving the operations:
[tex]\Delta y=50.4m[/tex]
Therefore, the displacement is 50.4 meters.
