Projectiles
Introduction
We will assume no air resistance in all problems.
Projectiles must be examined as two dimensional vectors. This motion
should be examined as having a horizontal and a vertical vector.
Always break the initial velocity into horizontal and vertical
components.
vo
horizontal = vo Cos
vo vertical = vo Sin
A key point to remember, there will be acceleration due to gravity in the
down direction. There is NO acceleration in the horizontal
direction.
There is a special case when the vertical displacement is 0. This
means the projectile ends at the same altitude it begins.
Assign Forward and Down as positive directions.
t = time in the air. Time is the only variable that is the same for
the horizontal and vertical directions.
Since the projectile hits the ground at the same altitude it was shot, x =
0.
vo vertical is negative and a is positive 9.80 m s -2
on Earth.
At its highest point, v vertical = 0
- vo vertical
v vertical = vo vertical + gt
= 0 => t =
-------------- since either vo or g is negative we
can cancel sign
g
The projectile is in the air twice the time it takes the projectile to
reach its highest point.
x horizontal = vo horizontal t
(Remember, no acceleration in horizontal direction)
Combining the equations.
2 vo horizontal vo vertical
2 vo2 Sin
Cos
vo2 Sin 2
x horizontal = -------------------------------- =
----------------------------- = -------------------- = Range
g
g
g
Sin 90o = 1 Thus the maximum range with no air resistance
is achieved when = 45o.
When solved for
g Range
Sin -1 (
-------------- )
vo2
=
----------------------------
2
The above equations can only be used when the projectile
lands at the same altitude it was shot.
Projectile hits the ground |
Calculate the time based only on vertical
information. Then use this time to calculate horizontal
displacement.
|
Projectile hits a wall |
Calculate the time based only on horizontal
information. Then use this time to calculate vertical displacement. |
Table of Contents
Projectile
Problems
1. |
A football is kicked off a tee with an
initial velocity of 20.0 m/s with an inclination of 30.0o.
How far downfield does the football hit the ground.
|
2. |
A soccer ball is kicked with an initial
speed of 23.0 m/s trying to hit a spot 50.0 m downfield. What angle
of inclination is necessary?
|
3. |
An arrow is shot from the top of a 200. m
cliff with an initial airspeed of 99.0 m/s and an inclination of 15.0o.
How far from the base of the cliff does the arrow hit the ground.
|
4. |
A cannon with an inclination of 25.0o
is 500. m from a canyon wall. If the initial airspeed is 150. m/s,
how high up the cliff does the cannon ball hit? |
Table of Contents
Projectile
Answers
1. |
A football is kicked off a tee with an
initial velocity of 20.0 m/s with an inclination of 30.0o.
How far downfield does the football hit the ground. How high does
the football get above the ground.
|
|
vo2 Sin 2
Range = -----------------
g
(20.0 m s -1)2 Sin (2 x 30.0o)
Range = ---------------------------------------- = 35.3 m
9.80 m s -2 |
The
ball hits at the same altitude it starts from allowing us to use the
Range Equation. |
|
vo
vertical = vo Sin
vo vertical = 20.0 m/s x Sin 30.0o
vo vertical = 10.0 m/s |
Calcule
the vertical component of the initial velocity. |
v
= vo + at = 0
-
vo - 10.0 m s
-1
t = ------- = ------------------- = 1.02 seconds
-
a - 9.80 m s -2
Y = vo vertical t + 1/2 gt2
Y = 10.0 m/s x 1.02 s + 1/2 x 9.80 m s -2
x (1.02 s)2
Y = 5.10 m |
The
highest point is reached when the vertical component to the velocity
is 0.
Up is positive and down is negative.
Use only the vertical information to find the highest point.
|
|
|
|
2. |
A soccer ball is kicked with an initial
speed of 23.0 m/s trying to hit a spot 50.0 m downfield. What angle
of inclination is necessary?
|
|
g Range
Sin -1 ( -------------- )
vo2 Sin 2
vo2
Range = ----------------- =>
= ----------------------------
g
2
9.80 m s -2 x 50.0 m
Sin -1
( ---------------------------- )
(23.0 m s -1 )2
= ------------------------------------------- =
33.9o
2 |
The ball hits at
the same altitude it starts from allowing us to use the Range
Equation. |
|
|
|
3. |
An arrow is shot from the top of a 200. m
cliff with an initial airspeed of 99.0 m/s and an inclination of 15.0o.
How far from the base of the cliff does the arrow hit the ground.
|
|
|
vo
horizontal = vo Cos
= 99.0 m/s x Cos 15.0o
= 95.6 m/s
vo vertical = vo
Sin
= 99.0 m/s x Sin 15.0o
= 25.6 m/s |
Break
the intial velocity into vertical and horizontal components. |
Down
is positive and up is negative
y = vo vertical t
+ 1/2 gt2
200. m = - 25.6 m/s t
+ 1/2 x 9.8 m/s2 t2
4.9 m/s2 t2
- 25.6 m/s t - 200. m = 0
t = - 4.29 s and
t = 9.51 s (negative
time not valid)
x = vo horizontal
t (No
Horizontal Acceleration)
x = 95.6 m/s x 9.51 s
= 909 meters |
Calculate
the time the arrow is in the air using the vertical information only
because the arrow hits the ground.
Solve for t using the quadratic
equation.
Use the time to calculate the horizontal
displacement. |
|
|
|
4. |
A cannon with an inclination of 25.0o
is 500. m from a canyon wall. If the initial airspeed is 150. m/s,
how high up the cliff does the cannon ball hit?
|
|
|
vo
horizontal = vo Cos
= 150. m/s x Cos 25.0o
= 136 m/s
vo vertical = vo
Sin
= 150. m/s x Sin 25.0o
= 63.4 m/s
|
Break the intial
velocity into vertical and horizontal components. |
x
= vo horizontal
t
(No Horizontal Acceleration)
x
500. m
t = ---------------- = ---------------
= 3.68 s
vo horizontal
136 m s -1
Up is positive and down is
negative
y = vo vertical
t + 1/2 gt2
y = 63.4 m s -1 x
3.68 s + 1/2 x ( - 9.80 m s -2 )(3.68 s)2
y = 167 m up the wall |
Calculate the
time the arrow is in the air using the horizontal information only
because the arrow hits the wall of the canyon.
Calculate the vertical displacement
using the time.
Acceleration is down. |
|
Table of Contents |