Mahela Plasencia - PhET exploring vectors and projectile motion

11. Was your prediction correct about the acceleration vector at this new angle? Yes 12.What did you notice about the acceleration vector at this new angle? the acceleration vector stayed the same throughout the entire flight of the cannonball 13.What was different about the vector (if anything) compared to the 45º or 60º angle? Move the cannon back to 45º or 60º if you need to check or verify anything. nothing the acceleration vector stayed the same throughout the entire flight of the two cannonballs 14.Summary: What have you discovered about the acceleration due to gravity of an object in flight with regards to the angle of launch? there is no difference in acceleration due to gravity for different angles Change the angle of the cannon to 90º. Fire the cannon at this new angle. Keep everything else the same. (Remember, click on the “slow” button to slow the simulation down)

Using the sin and cosine functions, calculate V x and V y. Show all work Vx=20 m/s (cos75) = 5.17 m/s Vy=20 m/s (sin75) = 19.31 m/s We will use the following formula to calculate the time to the top of the flight path: v f = v y + at Since at the top of the flight, the V f = 0, we have 0 = v y + at -v y = at t = - (V y / a) (Remember, a = -9.81 m/s 2 ) Using the bold formula above, acceleration due to gravity = -9.81 m/s 2 , and your v y that you calculated above, solve for (t). Show all work. -(19.31/-[9.8]) = 1.96 The value you just solved for is half of the time the ball is in the air. (Remember, it’s just to the top of the flight path). So take your time and double it. Show below: 3.92 Now, you solved for the total time in flight. Next, we’ll solve for distance in the x direction. We’ll use the following formula: v x = d x / t Rearranging for distance we get: d x = v x * t Using your v x that you calculated above, and the TOTAL flight time (the one you doubled), calculate the distance in the x direction. Show work below: (5.17)(3.92) = 20.26