0057 lecture notes - (part 1 of 2) an introductory projectile motion problem with … · 2021. 1....

0057 Lecture Notes - (Part 1 of 2) An Introductory Projectile Motion Problem with an Initial Horizontal Velocity.docx page 1 of 1 Flipping Physics Lecture Notes: An Introductory Projectile Motion Problem with an Initial Horizontal Velocity (Part 1 of 2) Example Problem: While in a car moving at 10.0 miles per hour, mr.p drops a ball from a height of 0.70 m above the top of a bucket. Part 1) How far in front of the bucket should he drop the ball such that the ball will land in the bucket? This is a projectile motion problem, so we should list our know variables in the x and y directions: x-direction: Δx = ?, v x = 10.0 mi hr × 1 hr 3600sec × 1609 m 1mi = 4.469 4 m s y-direction: Δy = −0.70 m, a y = −9.81 m s 2 , v iy = 0 Because we know three variables in the y direction, we should start there to find the change in time. Δy = v iy Δt + 1 2 a y Δt 2 ⇒−0.7 = 0 () Δt + 1 2 −9.81 ( ) Δt 2 ⇒ 2 () −0.7 ( ) = −9.81Δt 2 ⇒Δt 2 = 2 () −0.7 ( ) −9.81 ⇒Δt = 2 () −0.7 ( ) −9.81 = 0.377772sec Because change in time is a scalar and therefore independent of direction, we can use it in the x-direction: v x = Δx Δt ⇒Δx = v x Δt = 4.469 4 ( ) 0.377772 ( ) = 1.68843 ≈ 1.7m In the video I said, “If you decrease the speed of the car by half a mile per hour, you decrease the displacement in the x-direction of the ball by more than 8 cm.” Here is the mathematical proof: Let’s start by decreasing the car by half a mile per hour, this makes the initial velocity in the x-direction 9.5 mph. Known variables: x-direction: Δx = ?, v x = 9.5 mi hr × 1 hr 3600sec × 1609 m 1mi = 4.24597 2 m s y-direction: The y-direction information doesn’t change at all so we still get 0.377772 seconds for the change in time. (see, it didn’t change! ) Because change in time is a scalar and therefore independent of direction, we can use it in the x-direction: The difference in x displacement then is: (which is greater than 8 cm.) Δy = −0.70 m, a y = −9.81 m s 2 , v iy = 0 Δy = v iy Δt + 1 2 a y Δt 2 ⇒−0.7 = 0 () Δt + 1 2 −9.81 ( ) Δt 2 ⇒ 2 () −0.7 ( ) = −9.81Δt 2 ⇒Δt 2 = 2 () −0.7 ( ) −9.81 ⇒Δt = 2 () −0.7 ( ) −9.81 = 0.377772sec v x = Δx Δt ⇒Δx = v x Δt = 4.24597 2 ( ) 0.377772 ( ) = 1.60401m 1.68843 − 1.60401 = 0.084421m × 100cm 1m ≈ 8.4cm

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0057 Lecture Notes - (Part 1 of 2) An Introductory Projectile Motion Problem with an Initial Horizontal Velocity.docx page 1 of 1

Flipping Physics Lecture Notes: An Introductory Projectile Motion Problem

with an Initial Horizontal Velocity (Part 1 of 2)

Example Problem: While in a car moving at 10.0 miles per hour, mr.p drops a ball from a height of 0.70 m above the top of a bucket. Part 1) How far in front of the bucket should he drop the ball such that the ball will land in the bucket? This is a projectile motion problem, so we should list our know variables in the x and y directions:

x-direction: Δx = ?, vx =10.0mihr

× 1hr3600sec

× 1609m1mi

= 4.4694ms

y-direction: Δy = −0.70m, ay = −9.81ms2, viy = 0

Because we know three variables in the y direction, we should start there to find the change in time.

Δy = viyΔt +12ayΔt

2 ⇒ −0.7 = 0( )Δt + 12

−9.81( )Δt2 ⇒ 2( ) −0.7( ) = −9.81Δt2

⇒Δt2 = 2( ) −0.7( )−9.81

⇒Δt = 2( ) −0.7( )−9.81

= 0.377772sec

Because change in time is a scalar and therefore independent of direction, we can use it in the x-direction:

vx =ΔxΔt

⇒Δx = vxΔt = 4.4694( ) 0.377772( ) =1.68843 ≈1.7m In the video I said, “If you decrease the speed of the car by half a mile per hour, you decrease the displacement in the x-direction of the ball by more than 8 cm.” Here is the mathematical proof: Let’s start by decreasing the car by half a mile per hour, this makes the initial velocity in the x-direction 9.5 mph. Known variables:

x-direction: Δx = ?, vx = 9.5mihr

× 1hr3600sec

× 1609m1mi

= 4.245972 ms

y-direction:

The y-direction information doesn’t change at all so we still get 0.377772 seconds for the change in time.

(see, it didn’t change! J)

Because change in time is a scalar and therefore independent of direction, we can use it in the x-direction:

The difference in x displacement then is:

(which is greater than 8 cm.)

Δy = −0.70m, ay = −9.81ms2, viy = 0

Δy = viyΔt +12ayΔt

2 ⇒ −0.7 = 0( )Δt + 12

−9.81( )Δt2 ⇒ 2( ) −0.7( ) = −9.81Δt2

⇒Δt2 = 2( ) −0.7( )−9.81

⇒Δt = 2( ) −0.7( )−9.81

= 0.377772sec

vx =ΔxΔt

⇒Δx = vxΔt = 4.245972( ) 0.377772( ) =1.60401m

1.68843−1.60401= 0.084421m × 100cm1m

≈ 8.4cm

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