Y2, C6 - Projectiles

Question 1

What is the vertical acceleration with projectile motion

Answer 1

g

Question 2

What is the horizontal acceleration with projectile motion

Answer 2

0

Question 3

What does projectile motion being a parabola mean

Answer 3

They are symmetrical

Question 4

What can you consider separately with projectile motion

Answer 4

Vertical and horizontal components

Question 5

What value in suvat bridges vertical and horizontal motion

Answer 5

t - time
Time of the journey

Question 6

How do you find the distance travelled in projectile motion

Answer 6

Pythagoras of the horizontal and vertical distances

Question 7

What is an angle of:
a) elevation
b) depression

Answer 7

a) Angle upwards
b) Angle downwards

Question 8

What is the vertical velocity at the highest point of a projectile motion

Answer 8

0

Question 9

A particle is projected with speed 28m/s with angle of elevation 30 degrees, what are the horizontal and vertical components of velocity

Answer 9

Horizontal = 28 cos30
Vertical = 28 sin30

Question 10

If you take vertical motion downwards to be +ve, will your greatest height be +ve or -ve

Answer 10

-ve

Question 11

How do you find the time of a projectile motion journey

Answer 11

Set distance (s) to be 0 if projected from flat or the -ve of the elevation and solve suvat

Question 12

How do you find the length of time a particle is above a set distance

Answer 12

Set distance (s) to be the value
Solve simultaneously with suvat
Take the smallest t away from the biggest t to find answer

Question 13

How do you find the value for U (initial speed) when not given it

Answer 13

suvat for horizontal and vertical components, as t is equal you can solve for U as a numerical value

Question 14

Does horizontal speed change in projectile motion

Answer 14

NO

Question 15

What do tangents to the projectile motion tell you

Answer 15

The direction the particle is moving in

Question 16

A particle is projected from an elevated point A to a point B. How would you find the time after projection when the particle is moving parallel to AB

Answer 16

The tangent to the parabola would create a triangle
The triangle should be similar to the triangle OAB
Find gradient of triangle OAB
You know that the tangent triangle has a constant horizontal component so plug in the gradient and work out the vertical component from the gradient and horizontal component
Plug vertical velocity component into suvat to find time

Question 17

What is the range of a projectile motion

Answer 17

The horizontal distance it travels

Question 18

A particle is projected with angle x and velocity U, how would you find the range, R, on the horizontal plane

Answer 18

suvat the horizontal motion
u = Ucosx
a = 0
u = s/t –> s = ut
t = (2U * sinx) / g (derived from using vertical motion to find T)
rearrange suvat equations
R = (U^2 / g) * sin2x

Question 19

What equation for speed can you use with horizontal motion

Answer 19

speed = distance / time

Question 20

A particle is projected with angle x and velocity U, how would you find the time of the flight, T

Answer 20

suvat the vertical motion
u = Usinx
a = -g
s = 0
t = T
rearrange suvat and select the value of T where T doesn’t equal 0
T = (2U * sinx) / g

Question 21

For a particle with initial velocity U and at angle α above horizontal and moving freely under gravity:
What is the general result for time of flight

Answer 21

Time of flight = (2U sinα) / g

Question 22

For a particle with initial velocity U and at angle α above horizontal and moving freely under gravity:
What is the general result for time to reach greatest height

Answer 22

Time to reach greatest height = (U sinα) / g

Question 23

For a particle with initial velocity U and at angle α above horizontal and moving freely under gravity:
What is the general result for range on horizontal plane

Answer 23

Range on the horizontal plane = (U^2 sin2α) / g

Question 24

For a particle with initial velocity U and at angle α above horizontal and moving freely under gravity:
What is the general result for equation of trajectory

Answer 24

Equation of trajectory = x tanα - ((gx^2) / (2U^2)) * (1 + tan^2 α)