# Solving Projectile Motion Word Problems by cuiliqing

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```									   Unit Four: Projectile/Falling-Body
Motion 1
The first three formulas concern vertical motion;
the fourth concerns horizontal motion.
Formula 1: the distance a body falls from rest
because of gravity
y = ½ at2
Formula 2: the distance a body falls from gravity
with initial vertical velocity v0:
y = ½ at2 + v0t
Unit Four : Projectile/Falling-Body
Motion 2
Formula 3: the height of a body falling from gravity and
initial vertical velocity starting at an initial height, y0:
y = ½ at2 + v0t + y0
(y and y0 are in feet; t is in seconds; v0 is in feet/second;
a is −32 feet/sec.2; downward motion is negative)
Formula 4: (horizontal motion) the horizontal distance
traveled by a body moving with an initial horizontal
velocity (vH0) is as follows:
x = vH0t
(x is in feet; t is in seconds, and vH0 is in feet/second)
Unit Four: Projectile Motion –
Problem 1
Problem 1: Standing near the edge of the roof
of a building, Frank throws a baseball from a
height of 100 feet upwards with a vertical
velocity of 25 feet per second a height of 100
feet at a velocity. (As it falls, the ball will fall
past the roof down toward the ground.) How
high above the ground is the baseball after
three seconds?
Use the seven steps to solve this problem. If you
cannot, look at the solution shown below.
This is not really Frank . . .

. . . .but it’s who he’s pretending to be. (I know, Frank is
throwing with a vertical velocity, but he’s pretty
confused; don’t give him a hard time.)
Unit Four: Projectile Problem 1-
Solution I
•   Organize the data:
§    The initial height of the baseball is 100 feet
above the ground.
§    The vertical velocity is 25 feet/second.
§    The time elapsed is three seconds .
•   The problem asks for the height of the
baseball at that time (after 3 seconds)? That
variable is best expressed as y, which the
projectile formulas use as the vertical height.
Unit Four: Projectile Problem 1-
Solution II
4. We have three formulas from which to
choose for vertical height. This problem
involves not only the effect of gravity, but
also an initial velocity and an initial height,
given all of which we are to find the height at
a given time. Only Formula 3 seems to fit
this problem, so we choose it:
y = ½ at2 + v0t + y0
Unit Four: Projectile Problem 1- Solution
III
5.   Substitute the given data into the formula we have chosen:
y = ½ (−32)32 + 25(2) + 100
6. Solve mathematically:
y = ½(− 32)(9) + 50 + 100
y = − 144 + 150
y=6
Solution: after 3 seconds, the ball is at a height of 6 feet.
7. If you check the units of measurement in each term in the
formula, you will see that each separate term is in feet. In the
first term on the right of the equation, for example, feet/sec.2 is
multiplied by sec.2, so that the unit of measurement for the
entire term is feet.
Unit Four: Projectile Problem 2
Problem 2: Robin stands near the edge of a cliff
in the desert. She shoots an arrow into the air
from a height of 100 feet above the ground
below with a vertical velocity of 40 feet per
second a height and a horizontal velocity of 30
feet per second (sending it over the cliff).
What will be the height of the arrow relative to
the ground below the cliff when it has traveled
horizontally 75 feet?
Robin Practices!

“I shot an arrow into the
air. . . . . .
Unit Four: Projectile Problem 2 –
Solution I
•   We first read the problem until we
understand what tells and asks us.
•   We organize the problem data:
§   The object (arrow) begins at a height of 100 feet
relative to the ground below,
§   a vertical velocity of 40 feet per second,
§   a horizontal velocity of 30 feet per second,
§   and travels horizontally a total of 75 feet.
Unit Four: Projectile Problem 2 –
Solution II
3. We are asked what the height of the arrow is when
it has traveled 75 feet horizontally.
§   We are missing the amounts of two quantities: the length
of time the arrow travels to go 75 feet horizontally AND
the height of the arrow at that time.
§   We designate t as the length of time and y as the height of
the arrow (following the conventions in the Formulas).
4. We are asked to find the height, y, of the arrow.
Math Formula 3 seems most appropriate to find y
because we have both an initial height and an initial
vertical velocity. However, we cannot immediately
use Formula 3 because we do not yet know t, which
depends on the horizontal distance. To calculate t
first, we need to use Math Formula 4.
Unit Four: Projectile Problem 2 –
Solution III
5. & 6. Substep A. We substitute the data we
are given into Formula 4 and solve for t:
x = vH0t
75 feet = (30 feet/second)t
t = 2.5 seconds
Substep B. We take Formula 3, substitute the
given data and t = 2.5, and solve for y:
y = ½ at2 + v0t + y0
Unit Four: Projectile Problem 2 –
Solution IV
5. & 6. (cont.).
y = ½ at2 + v0t + y0
y = ½(−32ft/sec.2)(2.5sec.)2 + 40ft./sec.(2.5sec.)
+ 100 feet
y = (−16ft/sec.2)(6.25sec2) + 100 feet + 100 feet
y = −100 feet + 100 feet + 100 feet
y = 100 feet
The height is 100 feet above ground at that point.
7. We check the units which we calculated as
we proceeded, and they are correct.
Other Projectile/Falling-Body Problems
ØOther projectile/falling-body problems are
supplied in the accompanying written
materials for you to try. If you have difficulty,
guidance is available from tutors or in written
form.

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