10/18 Miley

Today’s class began with going over questions #1-16 in the Detailed Analysis packet. After this, Coats-Haan gave us the rest of class to complete the Detailed Analysis packet. During this time we were allowed to use our group members and the key as a reference on how to complete the problems. When most of the class reached problem #38, Coats-Haan gave us notes on the last kinematics equation. The last kinematics equation is range: R= V^2sin 2(theta)/g. In this equation V stands for speed and it can only be used if both Yf and Yi are the same. The numerical value for g is +9.8 m/s. After learning about the last equation, we were given time to continue the packet. It was at this time that Austin introduced his group members to his penguin eraser. If the Detailed Analysis packet wasn’t completed in class it became homework for the night.

QOD: To begin determining if the ball reaches the maximum height before or after it reaches the edge of the cliff you must determine the amount of time required for the ball to reach the top of its trajectory. This can be done using the 1^st kinematics equations which results in an answer of .587 sec. Then you must calculate the horizontal distance that the ball travels during the time it takes to reach the peak. This can be computed by using V= change in X/ change in T which gives you an answer of 5.85 m. The horizontal distance between the edge of the cliff and the initial X position is 5m (this is a given value). So to compute where the ball reaches its maximum height you can subtract 5.85-5 to give you an answer of .85. This means that the ball reaches its maximum height .85 m past the cliff.