14.5 mastery check: projectile motion\n1. a bean bag is launched up onto the top of a castle. the dots…

14.5 mastery check: projectile motion\n1. a bean bag is launched up onto the top of a castle. the dots represent the position of the bean bag at one - second intervals. ignore air resistance.\n4. identify the vertical velocity of the bean bag at the following points:\na. \nb. \nc. \nd. \ne. \nf. \ng. \n5. identify the horizontal velocity of the bean bag at the following points:\na. \nb. \nc. \nd. \ne. \nf. \ng. \n6. identify the acceleration of the bean bag at the following points:\na. \nb. \nc. \nd. \ne. \nf. \ng.\n\nat t = 0s, vx = 40 m/s, vy = 25 m/s

14.5 mastery check: projectile motion\n1. a bean bag is launched up onto the top of a castle. the dots represent the position of the bean bag at one - second intervals. ignore air resistance.\n4. identify the vertical velocity of the bean bag at the following points:\na. \nb. \nc. \nd. \ne. \nf. \ng. \n5. identify the horizontal velocity of the bean bag at the following points:\na. \nb. \nc. \nd. \ne. \nf. \ng. \n6. identify the acceleration of the bean bag at the following points:\na. \nb. \nc. \nd. \ne. \nf. \ng.\n\nat t = 0s, vx = 40 m/s, vy = 25 m/s

Answer

Explanation:

Step1: Recall projectile - motion concepts

In projectile motion, the horizontal acceleration (a_x = 0) (no horizontal force acting, ignoring air - resistance) and the vertical acceleration (a_y=-g=- 9.8\ m/s^{2}) (due to gravity). The horizontal velocity (v_x) is constant ((v_x = v_{0x})) and the vertical velocity (v_y) changes according to (v_y=v_{0y}-gt). Given (v_{0x}=40\ m/s) and (v_{0y} = 25\ m/s).

Step2: Find horizontal velocity at different points

Since there is no horizontal acceleration ((a_x = 0)), the horizontal velocity at all points A, B, C, D, E, F, G is (v_x=40\ m/s).

Step3: Find vertical velocity at different points

Use the formula (v_y = v_{0y}-gt). At (t = 0\ s) (point A): (v_{yA}=v_{0y}=25\ m/s) At (t = 1\ s) (point B): (v_{yB}=v_{0y}-g\times1=25 - 9.8\times1=15.2\ m/s) At (t = 2\ s) (point C): (v_{yC}=v_{0y}-g\times2=25-9.8\times2 = 5.4\ m/s) At the maximum - height (let's find the time to reach maximum - height (t_{max}=\frac{v_{0y}}{g}=\frac{25}{9.8}\approx2.55\ s)). Just before the maximum - height, the vertical velocity is still positive, and just after it is negative. At (t = 3\ s) (point D): (v_{yD}=v_{0y}-g\times3=25 - 9.8\times3=-4.4\ m/s) At (t = 4\ s) (point E): (v_{yE}=v_{0y}-g\times4=25 - 9.8\times4=-14.2\ m/s) At (t = 5\ s) (point F): (v_{yF}=v_{0y}-g\times5=25 - 9.8\times5=-24\ m/s) At (t) (point G): we can continue using the formula.

Step4: Find acceleration at different points

The acceleration in projectile motion is constant. The horizontal acceleration (a_x = 0\ m/s^{2}) and the vertical acceleration (a_y=-9.8\ m/s^{2}) at all points A, B, C, D, E, F, G.

At A: (v_{yA}=25\ m/s) At B: (v_{yB}=15.2\ m/s) At C: (v_{yC}=5.4\ m/s) At D: (v_{yD}=-4.4\ m/s) At E: (v_{yE}=-14.2\ m/s) At F: (v_{yF}=-24\ m/s) At G: Calculate using (v_y = v_{0y}-gt)

At A: (v_{xA}=40\ m/s) At B: (v_{xB}=40\ m/s) At C: (v_{xC}=40\ m/s) At D: (v_{xD}=40\ m/s) At E: (v_{xE}=40\ m/s) At F: (v_{xF}=40\ m/s) At G: (v_{xG}=40\ m/s)

At A: (a_{xA}=0\ m/s^{2},a_{yA}=-9.8\ m/s^{2}) At B: (a_{xB}=0\ m/s^{2},a_{yB}=-9.8\ m/s^{2}) At C: (a_{xC}=0\ m/s^{2},a_{yC}=-9.8\ m/s^{2}) At D: (a_{xD}=0\ m/s^{2},a_{yD}=-9.8\ m/s^{2}) At E: (a_{xE}=0\ m/s^{2},a_{yE}=-9.8\ m/s^{2}) At F: (a_{xF}=0\ m/s^{2},a_{yF}=-9.8\ m/s^{2}) At G: (a_{xG}=0\ m/s^{2},a_{yG}=-9.8\ m/s^{2})

Answer:

A: (v_y = 25\ m/s) B: (v_y = 15.2\ m/s) C: (v_y = 5.4\ m/s) D: (v_y=-4.4\ m/s) E: (v_y=-14.2\ m/s) F: (v_y=-24\ m/s) 5. A: (v_x = 40\ m/s) B: (v_x = 40\ m/s) C: (v_x = 40\ m/s) D: (v_x = 40\ m/s) E: (v_x = 40\ m/s) F: (v_x = 40\ m/s) 6. A: (a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}) B: (a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}) C: (a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}) D: (a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}) E: (a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2}) F: (a_x = 0\ m/s^{2},a_y=-9.8\ m/s^{2})