Lecture 25 - Projectile-Related Activities

Question 1

How can you determine the parabolic path of an airborne object?

Answer 1

• determined by the takeoff
• speed (how fast it is going) and direction (angle of release) at takeoff change what the parabola will look like

Question 2

what is the velocity vector?

Answer 2

• the direction at which the projectile is going
• can be split into horizontal and vertical vector components

Question 3

what is the vertical vector component?

Answer 3

• how high the projectile is going
• affected by gravity

Question 4

what is the horizontal vector component?

Answer 4

• how far the projectile is going
• not affected by gravity
• stays the same the whole time

Question 5

when is the vertical vector equal to zero?

Answer 5

• at the peak of the parabola (R=H)
• where R = changes in resultant

Question 6

what is the best trade/ maximum horizontal distance?

Answer 6

• throwing at 45 degrees

Question 7

what is air resistance?

Answer 7

• the drag force to the projectile related to the speed squared
• how much force is required for the air to split and reform after the projectile
• affects both horizontal and vertical velocity

Question 8

what would the projectile parabola look like if air resistance was not present?

Answer 8

• would be symmetrical (if started and landed on a surface at the same level)

Question 9

how can you calculate the max height of a projectile when air resistance is not present?

Answer 9

• height = (v vert)^2 / 2g
• where v vert = vertical component of takeoff velocity
• where g = acceleration due to gravity

Question 10

how can you maximize height if the projectile is your body?

Answer 10

• “rearrange” body segments to get our center of mass higher
• ## the lower on your body the center of mass is, the higher you can reach (because the center of mass will rise the same amount)

Question 11

how is height maximized in pole vaulting?

Answer 11

• the bending of the pole converts horizontal speed to vertical speed (this is a special case)

Question 12

how can you maximize horizontal distance?

Answer 12

• distance = v horizon x t
• where v horizon = horizontal component of takeoff velocity
• where t = amount of time spent in the air
• 2 angles will result in the same horizontal trajectory (one more vertical throw and one more horizontal throw) except for 45 degrees which is peak

Question 13

how can you find t?

Answer 13

• found from the vertical component (calculate peak)

Question 14

how does height affect max horizontal distance?

Answer 14

• for the same horizontal takeoff component, starting higher will give a greater horizontal distance

Question 15

how does speed affect max horizontal distance?

Answer 15

• for the same takeoff height, a faster horizontal component will give a larger horizontal distance

Question 16

how does angle affect max horizontal distance?

Answer 16

• with too high a takeoff angle, with the same takeoff velocity, you can shorten the distance travelled

Question 17

how do long jumpers maximize horizontal distance travelled?

Answer 17

• take off between 18 and 23 degrees
• too high = velocity is wasted on height
• too flat = not in the air long enough
• the center of mass stays below the parabola
• feet in front for the landing to maximize distance (but not too far in front, they don’t want to fall on their bum)

Question 18

how are accuracy and margin of error affected by horizontal distance/angle?

Answer 18

• the approach direction changes the margin of error (straight on = a circle, to the side = an oval)
• maximize the margin of error by standing directly in front of the dartboard