Part 2:projectiles launched at an angle

Pages 102 – 104

Motion in Two Dimensions

Maximum height and range • When objects are launched at an angle, there

is a component of their velocity in the horizontal direction and a component in the vertical direction.

• The horizontal velocity would be vicosθ

• The vertical velocity would be visinθ

• Problems can then be worked in the same way.

• Keep in mind that the vertical velocity at the peak of the path is zero.

Launched at an angle

Velocity

•Final speed = initial speed (conservation of energy)

•Impact angle = - launch angle (symmetry of parabola)

Launched at an angle•The launch angle has an affect on

the range of the projectile.

•As the angle increases from 0 to 45 degrees, the range increases.

•At 45 degrees, the range is at a maximum

•As the angle increases from 45 degrees to 90 degrees, the range decreases.

Launched at an angle

Launched at an angle

0

5

10

15

20

25

30

35

0 20 40 60 80

15 deg

30 deg

45 deg

60 deg

75 deg

Equations of motion:

X

Uniform motion

Y

Accelerated motion

ACCELERATION ax = 0 ay = -9.81 m/s2

VELOCITY vx = vi cos θ vyf = vi sin θ + aΔt

vfy2 = vi

2sin θ +2a Δy

DISPLACEMENT Δx = vi cos θ Δt Δy = visinθΔt + ½ aΔt2

The Hulk throws a boulder onto a police car trying to arrest him. The initial velocity of the boulder is 12 m/s and he throws it at an angle of 39 degrees to the horizontal:

a. Find the horizontal and vertical components of the initial velocity

b. Find the maximum height of the boulder

c. Find the time the boulder was in the air

d. Find the horizontal distance the boulder traveled

39°

vi

vi = 12m/s

39°

vi

Givensangle = 39°vi = 12m/sa = -9.8 m/s2

Equationsvx = vi cos θvy = vi sin θ

Δx = vi cos θ Δt

vyf = vi sin θ + aΔt

Δy = visinθΔt + ½ aΔt2

vfy2 = vi

2sin θ +2a Δy

Finding the componentsvxi = vi cos θ

vyi = vi sin θ

Givensangle = 39°vi = 12m/sa = -9.8 m/s2

Equationsvx = vi cos θvy = vi sin θ

Δx = vi cos θ Δt

vyf = vi sin θ + aΔt

Δy = visinθΔt + ½ aΔt2

vfy2 = vi

2sin θ +2a Δy

Find time first!!!

vyf = vi sin θ + aΔt

Givensangle = 39°vi = 12m/sa = -9.8 m/s2

Equationsvx = vi cos θvy = vi sin θ

Δx = vi cos θ Δt

vyf = vi sin θ + aΔt

Δy = visinθΔt + ½ aΔt2

vfy2 = vi

2sin θ +2a Δy

Finding max height

Δy = visinθΔt + ½ aΔt2

Givensangle = 39°vi = 12m/sa = -9.8 m/s2

Equationsvx = vi cos θvy = vi sin θ

Δx = vi cos θ Δt

vyf = vi sin θ + aΔt

Δy = visinθΔt + ½ aΔt2

vfy2 = vi

2sin θ +2a Δy

Finding the time in the air Givensangle = 39°vi = 12m/sa = -9.8 m/s2

Equationsvx = vi cos θvy = vi sin θ

Δx = vi cos θ Δt

vyf = vi sin θ + aΔt

Δy = visinθΔt + ½ aΔt2

vfy2 = vi

2sin θ +2a Δy

Finding the horizontal distance

Δx = vi cos θ Δt

Givensangle = 39°vi = 12m/sa = -9.8 m/s2

Equationsvx = vi cos θvy = vi sin θ

Δx = vi cos θ Δt

vyf = vi sin θ + aΔt

Δy = visinθΔt + ½ aΔt2

vfy2 = vi

2sin θ +2a Δy