Projectile Motion

Free Fall - Word Problems

Worked Examples: Displacement, Velocity, Time, and Acceleration

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Do Now warm-up prompt

B. Free Fall - falling where the only force is gravity (no friction)

Why would a crumpled piece of paper dropped from a height of 3 m have a greater acceleration than an uncrumpled one?

What Is a Vacuum?

Brain Bites - NASA

School blocks YouTube? Download the file below.
vacuum_hi.mov

Video Instruction

showmethephysics.com

Slideshow

1. Free Fall Word Problems

Use the same method and equations as before, except always remember when you read…

Step 1 - Read and underline all the important information.

Step 2 - Turn all your information into letters.

Step 3 - Substitute and solve.

“…an object is dropped”

A raindrop falling

Vi = 0

a = g = +9.8 m/s2

(make down +)

“…an object is ascending”

A monkey jumping up and down

a = g = -9.8 m/s2

(make down -)

Try It

Object thrown up, Vi = 110 m/s.
Sketch a V vs. t plot of the object from the throw to the peak (use a = -10 m/s2).

Blank V vs. t graph for sketching
Kinematics equation reference chart

a. Simple Free-Fall Problems

The Leaning Tower of Pisa

Leaning Tower of Pisa, in Pisa, Italy.

(Wednesday 19 May 2004, 17:37)

The lean of this tower has been tried to fix several times. The most recent time I believe was in 2001 when they tried to drill into the ground and pour cement underneath the sandy surface.

Ex

Given

A coin is dropped in a vacuum tube.
Find: its displacement after .30 seconds.

↓

Vi = 0, a = +9.8 m/s2 (dropped), d = ?, t = .30 sec

d = Vit + ½at2 = 0(.30) + ½(9.8 m/s2)(.30 s)2

d = 0 + 4.9(.09)

d = .44 m

Ex

Given

A coin is dropped in a vacuum tube.
Find: the coin's velocity after .30 seconds.

↓

Vi = 0 m/s, a = +9.8 m/s2 (dropped), Vf = ?, t = .30 sec

Vf = Vi + at   OR   a = (Vf - Vi)/t

Vf = 0 + 9.8 m/s2(.3 sec)

Vf = 2.9 m/s

Ex

A rocket is fired vertically upward with an initial velocity of 29 m/s.

[ Flash ]

Given

Find: the rocket's maximum altitude.

↓

At Maximum Altitude, Vf = 0 m/s

Vi = +29 m/s, Vf = 0 m/s, a = g = -9.8 m/s2, d = ?

Vf2 = Vi2 + 2ad

(0 m/s)2 = (+29 m/s)2 + 2(-9.8 m/s2)d

0 = 841 + -19.6d → -841 = -19.6d

d = 43 m

Stroboscopic photo of a falling ball, BigDropMotionShot
Uniform or Non-uniform Acceleration?

Ex

Given

An object dropped on planet X falls 64 m in 4 seconds.
Find: the acceleration due to gravity on planet X.

↓

Vi = 0 m/s, d = 64 m, t = 4 sec, a = ?

d = Vit + ½at2 → 64 m = 0 m/s(4 sec) + ½a(4 sec)2

64 m = 0 + 8a

a = 8 m/s2

AP Question

Diagram of a ball thrown upward from a height above the ground
Given

A ball is thrown upward from a position (y0) above the ground with a velocity of 8 m/s.
Find: y0, if it took 3.0 seconds for the ball to hit the ground.

↓

ay = -10. m/s2 (g), yi = ?, Vi = 8 m/s, yf = 0 (hits the ground), t = 3.0 sec

Δd = Vit + ½at2 → yf - yi = Vit + ½at2

0 = yi + 8 m/s(3.0 sec) + ½(-10. m/s2)(3.0 sec)2

0 = yi + 24 m + -45 m → -yi = -21 m

yi = 21 m

Given

Same ball as above.
Find: what velocity does the ball hit the ground?

↓

Make down positive - treat this as a ball falling from a height of 21 m with a velocity of 8 m/s down.

ay = +10. m/s2 (g), Δd = +21 m, Vi = +8.0 m/s (velocity becomes +8 m/s when the ball returns to its original position), Vf = ?

Vf2 = Vi2 + 2ad = 64 + 2(10 m/s2)(21 m)

OR: Vf = 8.0 m/s + -10 m/s2(3 sec) = -22 m/s

Vf = 22 m/s down

Conceptual Check

Paul Hewitt Question

When you drop a ball it accelerates downward at 9.8 m/s2. If you instead throw it downward, then its acceleration immediately after leaving your hand, assuming no air resistance, is…

A.  9.8 m/s2.
B.  more than 9.8 m/s2.
C.  less than 9.8 m/s2.
D.  Cannot say, unless the speed of throw is given.

Ex

Given

A basketball player jumped straight up to grab a rebound. She was in the air for 0.80 second.
Find: how high did she jump?

↓

Treat this as the player coming down from the peak of her rebound (half the hang time).

ay = 10. m/s2 (g), d = ?, Vi = 0 m/s, t = .40 sec

Δd = Vit + ½at2 = ½(10 m/s2)(.4 sec)2

Δd = .80 m

Ex

Given

A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0 m high.
Find (a): how long does it take to reach the bottom of the cliff?

↓

t = ? (how long), ay = 9.80 m/s2 (down positive), Viy = -12.0 m/s, Δdy = +70.0 m (down positive)

Δdy = Viyt + ½ayt2

70.0 m = -12.0 m/s(t) + 4.9 m/s2(t2)

0 = -12.0 m/s(t) + 4.9 m/s2(t2) - 70 m   (quadratic, positive root)

t = 5.20 sec

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