2 dimensional motion

• Let’s review what we already know...
• 2 dimensional motion is simply the combination of motion in the x direction with motion in the y direction

• The curved (parabolic) shape of this motion is the natural result of this combination
• The general motion equation holds: s_f = s_i + v_it + ¹/₂at²

• We get 2 versions of the equations - x & y direction

x direction: s_fx= s_ix + v_ixt + ¹/₂a_xt²

x direction: s_fy= s_iy+ v_it + ¹/₂a_yt²

Horizontally projected

a_x = 0 and s_fx= s_ix + v_ixt

• If it has a rocket motor, a_x>0
• If we include air resistance, a_x<0.

• For a horizontally projected object, there is no initial vertical velocity, so s_fy= s_iy– 4.9 t²
• The object moves forward at a constant speed only while it is airborne!
• Time is the same in both equations

Steps to Solution

• You will be given information to solve either the x-direction or y-direction for time
• Then use the time to solve for the unknown direction
• Usually you solve for the y-direction first

Sample - Thelma & Louise

• A car is driven horizontally off a 100 m cliff at 25 m/s. How far from the base of the cliff will the wreckage be found?

• Solve y-direction first to find time

• -100m=(0m/s)t+0.5(-9.8m/s²)t²
• (-100 m)/(-4.9m/s²) = t²
• 20.4 s² = t²
• t² = 20.4 s²
• t = 4.5 s
• Then solve x direction using time found

• s_x = 25 m/s (4.5s)
• s_x = 112.5 m

Parametric equations

• These results could also be found by graphing
• recall: s = s_f - s_i
• So s_fx=s_ix+ v_ixt + ½a_xt²
• And s_fy=s_iy+ v_iyt + ½gt²
• Using the parametric mode:
• X_T1 = 0 + 25T + 0.5(0)T²
• Y_T1 = 100 + 0T + 0.5(-9.8)T²

Practice Problem

• A ball is thrown horizontally off a 20 m high tower at 15 m/s.
• How long is it in the air?
• Where does it hit the ground?
• What is the magnitude of its velocity when it hits the ground?

Answer:

• t = 2.02 s, s_x = 30.3 m
• First, pick out known & unknown values from the problem.

• Solve y-direction first to find time

• -20m=(0m/s)t+0.5(-9.8m/s²)t²
• (-20 m)/(-4.9m/s²) = t²
• 4.04 s² = t²
• t² = 4.04 s²
• t = 2.02 s
• Then solve x direction using time found

• s_x = 15 m/s (2.02s)
• s_x = 30.3 m
• The same thing can be done with the velocity equation, x and y are still independent: v_f = v_i + at
• So in the x direction:
• v_fX = v_iX + a_Xt
• a_x=0 so v_fx= v_ix = 15 m/s

• And in the y direction:
• v_fY = v_iY + a_Yt
• v_fy = 0m/s + (-9.8 m/s²)(2.02s)
• v_fy = -19.8 m/s
• Then combine the vectors!
• v_fx= 15 m/s v_fY = -19.8 m/s
• Use Pythagorean theorem to get the magnitude and & inverse tangent to get direction.

• v_f = 24.8 m/s, 307degrees

Projected at an angle

2-D Motion Equations

• These same equations we used in the first case (horizontally projected) hold true when the object is initially thrown at an angle.
• BUT Initial velocity in the y direction is NOT zero, so it does NOT cancel out!

2 - d projected at an angle

• The initial velocity must be broken down into an x and y component using right triangle trigonometry
• Then the problem is solved like before.
• v_ix = v_i cos q
• v_iy = v_i sin q

Note:

• The path of our (Case 2) projectile is a symmetric parabola.
• Speed at takeoff = speed at landing.
• Angles are equal in magnitude, opposite in direction.

Sample Problem

• The six million dollar man kicks a ball at a 30 degree angle giving it a velocity of 100 m/s.
• How long is it in the air?
• How far away does it land?
• How fast is it going when it hits the ground?
• How high does it rise?

Gather x & y variables

• Step 1- resolve the velocity into x & y components

Hang time:
How long is it in the air?

• Before it is kicked and when it hits the ground, its height above the ground is zero so its change in height s_y = 0.
• s_y = v_iy D t + 0.5 g D t²
• 0 = 50 D t - 4.9 D t²
• 0 = D t (50 - 4.9 D t)
• 0 = D t OR
• 0 = 50 - 4.9 D t
• 50 = 4.9 D t
• D t = 10.2 s

Range:
How far away does it land?

• s_x = v_ix D t
• s_x = 86.6 m/s (10.2s)
• s_x = 884 m

Velocity:How fast is it going?

• Find v_fx& v_fy @ t=1.02 s using: v_f = v_i + at
• v_fx = v_ix + a_xt v_fy = v_iy + a_yt
• v_fx = 86.6m/s + 0t = 86.6 m/s
• v_fy = 50 m/s +(-9.8m/s²⁾(10.2s)
• = -49.96 m/s
• Use Pythagorean theorem & trig
• V_f = ((86.6)² + (-49.96)²)^½ = 99.98 m/s = ~ 100 m/s
• q = tan^-1 (-4.996/8.66) = -30º in quadrant IV v_f = 100 m/s, 330º

Max height

• Just like the case of a ball thrown vertically upward,
• Max height occurs when v_fy = 0.
• V_fy² = v_iy² + 2a_ys_y
• 0² = 50² + 2(-9.8)s_y
• -2500 = -19.6s_y
• s_y = 128 m

Monkey and Zookeeper

Practice Problem 2

• A ball is kicked at a 60 degree angle at 20 m/s.
• How long is it in the air?
• Where does it hit the ground?
• How high is it at its highest point?
• What is the magnitude of its velocity when it hits the ground?

Variables

• Resolve the velocity into components
• v_ix = v_i cos q = 20 cos 60 = 10 m/s
• v_iy = v_i sin q = 20 sin 60 = 17.3 m/s

Solve y direction

• Before it is kicked and when it hits the ground, its height above the ground is zero so its change in height s_y = 0.
• s_y = v_iy D t + 0.5 g D t²
• 0 = 17.3 D t - 4.9 D t²
• 0 = D t (17.3 - 4.9 D t)
• 0 = D t 0 = 17.3 - 4.9 D t
• 17.3 = 4.9 D t
• D t = 3.53 s

Solve x direction

• s_x = v_ix D t
• s_x = 10 m/s (3.53s)
• s_x = 35.3 m

Max height

• Max height occurs when v_fy = 0.
• v_fy² = v_iy² + 2 g s_y
• 0 = (17.3)² - 19.6 s_y
• 19.6 s_y = 299.3
• s_y = 15.3 m

Solve velocity

• Find v_fx & v_fy @ t = 3.53 s using
• v_f = v_i + at and then use
• Use Pythagorean theorem & trigonometry to find v_f
• OR
• APPLY symmetry
• V_f = v_iy= 20 m/s
• q _f = - q _i = -60º in quadrant IV
• so v_f = 20 m/s, 300º

Projected at angle from a height

Initial displacement

• If the object is initially some height above the ground when it is projected at an angle, it is usually necessary to use the quadratic formula. It will not easily factor!
• If its initial height is 0, then you can factor out a t!

Sample Problem

• A ball is kicked off a 50 m tower at a 30 degree angle giving it a velocity of 10 m/s.
• How long is it in the air?
• How far away does it land?
• At what velocity does it hit the ground?
• How high does it rise?

Solve

• Step 1- resolve the velocity into x & y components
• v_ix = v_i cos q = 10 cos 30 = 8.66 m/s
• v_iy = v_i sin q = 10 sin 30 = 5 m/s

Find time in air

• Its height above the ground is 50m so its change in height s_y = -50m.
• D s_y = v_iy D t + 0.5 g D t²
• -50 = 5 D t - 4.9 D t²
• 4.9t² - 5t - 50 = 0
• a = 4.9 b = -5 c = -50
• Use quadratic formula - discard negative answer
• t = (-b ± Ö b² - 4ac) / (2a)
• t = (-(-5) ± Ö (-5)² - 4(4.9)(-50) ) / (2(4.9))
• t = - 2.7, 3.7s so t = 3.7s when it hits the ground

How far away does it land?

• s_x = v_ix D t
• s_x = 8.66 m/s (3.7s)
• s_x = 32.04 m

Velocity:How fast is it going?

• Find v_fx& v_fy @ t=3.7 s using: v_f = v_i + at
• v_fx = v_ix + a_xt v_fy = v_iy + a_yt
• v_fx = 8.66m/s + 0t = 8.66 m/s
• v_fy = 5 m/s +(-9.8m/s²⁾(3.7s)
• = -31.26 m/s
• Use Pythagorean theorem & trig
• V_f = ((8.66)² + (-31.26)²)^½ = 32.4 m/s
• q = tan^-1 (-31.26/8.66) = -74.5º in quadrant IV v_f = 32.4 m/s, 285.5º

Max height

• Just like the case of a ball thrown vertically upward,
• Max height occurs when v_fy = 0.
• V_fy² = v_iy² + 2a_ys_y
• 0² = 5² + 2(-9.8)s_y
• -25 = -19.6s_y
• s_y = 1.28 m above starting point OR 50 m + 1.28 m = 51.28 m above the ground

Practice Problem

• A ball is projected off a 120 m high tower at a 50 degree angle at 25 m/s.
• How long is it in the air?
• Where does it hit the ground?
• What is the magnitude of its velocity when it hits the ground?

Solutions

• t = -3.4, 7.3 s so t = 7.3 s when it hits the ground
• s_x = 117.3 m
• v_f = 54.8 m/s, 290º
• s_y = 18.7 m above starting point OR 120 m + 18.7 m = 138.7 m above the ground

Solve

• Step 1- resolve the velocity into x & y components
• v_ix = v_i cos q = 25 cos 50^o = 16.07 m/s
• v_iy = v_i sin q = 25 sin 50^o = 19.15 m/s

Find time in air

• Its height above the ground is 50m so its change in height s_y = -50m.
• D s_y = v_iy D t + 0.5 g D t²
• -120 = 19.15 D t - 4.9 D t²
• 4.9t² - 19.15t - 120 = 0
• a = 4.9 b = -19.15 c = -120
• Use quadratic formula - discard negative answer
• t = (-b ± Ö b² - 4ac) / (2a)
• t=(-(-19.15)± Ö (-19.15)²-4(4.9)(-120))/(2(4.9))
• t = -3.4, 7.3 s so t = 7.3 s when it hits the ground

How far away does it land?

• s_x = v_ix D t
• s_x = 16.07 m/s (7.3s)
• s_x = 117.3 m

Velocity:How fast is it going?

• Find v_fx& v_fy @ t=3.7 s using: v_f = v_i + at
• v_fx = v_ix + a_xt v_fy = v_iy + a_yt
• v_fx = 16.07m/s + 0t = 16.07 m/s
• v_fy = 19.15 m/s +(-9.8m/s²⁾(7.3s)
• = -52.39 m/s
• Use Pythagorean theorem & trig
• V_f = ((19.15)² + (-52.39)²)^½ = 54.8 m/s
• q = tan^-1 (-52.39/19.15) = -70º in quadrant IV v_f = 54.8 m/s, 290º

Max height

• Just like the case of a ball thrown vertically upward,
• Max height occurs when v_fy = 0.
• V_fy² = v_iy² + 2a_ys_y
• 0² = 19.15² + 2(-9.8)s_y
• -366.7 = -19.6s_y
• s_y = 18.7 m above starting point OR 120 m + 18.7 m = 138.7 m above the ground