I’m going to use the equation     Vf = Vi + at      after we rearrange    t = (Vf – Vi) ÷ a        t = (-11.5 m/s – 11.5 m/s) ÷ -9.8 m/s2 =   2.35 seconds With a time of 2.35 m/s, and using the equation for constant velocity for our horizontal range, we calculate the range      Range = Horizontal Velocity x time            Range   =   9.6 m/s x 2.35 sec   =   22.6m Question: Now, set the muzzle velocity of the cannon to 12 m/s and set the angle of the cannon to 300. First calculate the time and the range using the above approach

3. The graph below shows the displacement vs time of a student leaving the NSCI 110 after physics class walking towards the cafeteria for lunch (assume a 1-D motion) At some point between NSCI 110 and the cafeteria, the student realizes they have left their cell phone in the classroom, so they turn back to get it. Displacement vs Time 120 100 nol Velocity vs Time 80 1.5 60 40 30 60 90 120 150 180 210 20 -1.5 30 60 90 120 150 180 210 Time (s) a. How far did the student walk before realizing that they left their cell phone in the classroom? How long did this take? (2 marks) b. With what velocity did the student walk back to the classroom? (2 marks) c. Fill out the velocity vs time graph above. (2 marks) Distance (m)

A ball, rolling with velocity 2 m/s, rolls off a cliff. The cliff ledge is 90 m above the ground below. What is the horizontal distance the ball will travel before hitting the ground. Give numerical answer in meters.

A cannon was launched with an initial speed of 15 m/s. How do you calculate the horizontal (V0x)  and the vertical  (V0y) components of the initial velocity, V0, for each time? Given this info: Time       Range (x)     Height (y)   V0x.        V0y 0.             0                    0                 ?                ? .1            1.3                   .7                ?                ? .2           2.6                  1.3                ?              ?