Projectiles

Question 1

What are projectiles?

Answer 1

Objects that move in the air in 2 dimensions (not just straight up or down)

Question 2

What are the 4 assumptions when working with projectiles?

Answer 2

• Object is modelled as a particle
• weight is the only acting force so air resistance, lift etc are ignored
• object begins flight with specific initial velocity
• spin and dimnsions of object are ignored

Question 3

How do you solve horizontal projection problems?

Answer 3

resolve into horizontal and vertical components, identifying a suvat list for each. Horizontal acceleration is always 0 and vertical is always -9.8 thats it.

Question 4

how do you deal with projection at an angle (parabola)

Answer 4

assume ground is flat and level. Accelerations are the sam as with horizontal projectiles but velocity up and across must be resolved using trig to begin. at top of parabola, when time is 1/2 t and horizontal dist = 1/2 s, v vertically = 0, v horizontally is constant. here vertical disp is greatest.

Question 5

how tackle projections at an angle above ground?

Answer 5

carefully using same methods as with projections from ground but with drawing diagrams

Question 6

how do you find time for which an object is above a certain height?

Answer 6

find time to reach height, time to reach max height, subtract and double

Question 7

how do you do the vector method with projectiles?

Answer 7

write out suvat and organise vectors into the variables, then simply place vectors into required suvat equations and work out the horizontal ond vertical components on the two lines. deal with magnitudes using pythagoras where applicable

Question 8

what are the projectile formulae?

Answer 8

a set of formulae involving trig functions that are not to be learned but derrived in the exam, its worth being vaguely familliar with them.

Question 9

what are the projectile formulae for time of flight, time to reach greatest height, range on horizontal and equation of trajectory?

DO NOT LEARN.

Answer 9

time of flight: (2U sin a) / g
time to reach greatest height:
(U sin a) / g
range on horizontal: (U^2 sin2a) / g
equation of trajectory:
y = x tan a - (gx^2/2U^2)(1+tan^2 a)
where y is vert. height and x is horiz. dist.