University:
Montgomery College
Course:
PHYS 161: General Physics I: Mechanics and Heat
Academic year:

2019
Views:

499

Pages:

2
Author:

96XGbl6o

Simple Projectile Motion: In simple projectile motion, the only force is Due to gravity Constant in magnitude (g = 9.8 m/s2) Vertical We can separate the motion of an object into horizontal and vertical components: Horizontal: Motion with constant velocity ax where xi yi ay 0 vxf xf Vertical: Motion with constant acceleration vyf vxi xi yf vxi t 9.8 m⁄s g, g vyi yi gt vyi t gt and xf yf are initial and final positions, vi and vf are initial and final speeds, and t is time (ti = 0). Maximum height vy t up 0 +Y t down +X vi Ө In the figure, a projectile is launched with initial speed vi at angle Ө . So the initial vertical speed vyi vi sin Ө and the initial horizontal speed vxi vi cos Ө Tips: Once launched, the projectile follows a parabolic path. At the maximum height, the vertical speed will be zero (otherwise the projectile would go higher). It takes the same time to reach the peak from any height ( t up ) as it does to fall from the peak to that same height ( t down ). 3 = Maximum height: vy 2 0 +Y 4 5 1 6 vi Ө vyi +X vxi vi cos Ө vi sin Ө 7 Tips (continued): Because parabolas are symmetric, when the projectile is at the same heights, the speed going up will be the same as the speed going down. The figure shows that the vertical speeds at positions 1 and 5 are the same, as are the vertical speeds for positions 2 and 4. To find the maximum height, solve for the time when the vertical speed is zero. Then use that time to find the distance traveled. It’s often easier to break a problem into segments. For example, the time required to go from position 1 in the figure to e.g., position 7 may be broken up into 2 calculations: 1. Calculate tup , the time needed for the projectile to move from 1 to 3. 2. Calculate tdown , the time needed for the projectile to move from 3 to 7. tdown will be given by yi yf g t down 3. Add tup + tdown to get the total time needed. However, to see whether the projectile will clear an obstacle (e.g., figure position 6): 1. Calculate t6 , the time needed to travel the horizontal distance (e.g., from 1 to 6). t6 will be given by xf xi vxi t 2. Then calculate the projectile height at t6 ,and compare it to the height of the obstacle. Remember that sometimes it’s easiest to use the formula vyf vyi 2 g yi yf