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Exp. 7: vary the angle to maximize the height, Purpose, Theory – PASCO ME-6800 Projectile Launcher (Short Range) User Manual

Page 35

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M o d e l N o . M E - 6 8 0 0

E x p . 7 : V a r y t h e A n g l e t o M a x i m i z e t h e H e i g h t

0 1 2 - 0 5 0 4 3 G

31

Exp. 7: Vary the Angle to Maximize the Height

Equipment Needed

Purpose

The purpose of this experiment is to find the launch angle that will maximize the height on a vertical wall for a
projectile launched at a fixed horizontal distance from the wall.

Theory

When the ball is shot at an angle at a fixed distance, x, from a
target such as a vertical wall, the ball hits the wall at a height y
given by:

where y

0

is the initial height of the ball, v

0

is the initial speed

of the ball as it leaves the muzzle,

 is the angle of inclination

above horizontal, g is the acceleration due to gravity, and t is
the time of flight. The range is the horizontal distance, x,
between the muzzle of the Launcher and the place where the
ball hits, given by

Solving this equation for the time of flight, t, gives

Substituting for t in the equation for y gives

To find the angle,

, that gives the maximum height, y, find the first derivative of the equation for y and set it equal

to zero. Solve for the angle,

.

Solving for the angle,

, gives:

Since the second derivative is negative for

max

, the angle is a maximum. To find the inital speed of the ball, use

the fixed distance, x, and the maximum height, y

max

. Solve the y-equation for v

0

and plug in the values for y

max

,

max

, and x.

Item

Item

Projectile Launcher and plastic ball

Board to protect wall

Meter stick or measuring tape

Sticky tape

White paper, large sheet

Carbon paper (several sheets)

Plumb bob and string

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PROJECTILE LAUNCHER

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in

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90

80

70

60

50

40

30

20

10

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θ

y

0

x

y

υ

0

Figure 7.1: Maximizing Height

Range

Initial

height

Angle

Initial

speed

y

y

0

v

0

sin

t 1

2

---gt

2

+

=

x

v

0

cos

t

=

t

x

v

0

cos

-----------------

=

y

y

0

x

gx

2

2v

o

2

2

cos

-----------------------

tan

+

=

dy

d

------

x

2

gx

2

2

sec

tan

v

0

2

----------------------------------

sec

0

=

=

max

tan

v

0

gx

------

2

=

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