Game Development Reference
In-Depth Information
Answers to Exercises
1.
The time of impact is the time when the vertical or z-position of the projectile will be
zero. The time of impact can be determined from Equation (5.8c).
2
z
v
1
t
=
0
−
gt
2
+
v
t
=
0
so
z
g
0
2
The z-component of velocity at the time of impact can be found from Equation (5.8b).
2
z
v
v
=−
gt
+
v
=−
g
0
+
v
=−
v
z
z
0
z
0
z
0
g
The x-component of velocity using the gravity-only model is constant throughout the
trajectory. Therefore, the velocity of the projectile at impact is equal to the following:
2
2
2
2
(
)
v
=+= +− =
v
v
v
v
v
x
z
xo
z
0
0
The velocity at impact under the gravity-only model is equal to the initial velocity of the
projectile.
2.
The golf ball will continue to travel through the air until gravity brings it back down to
the ground. Under the gravity-only model, the vertical location of the ball as a function
of time is given by Equation (5.8c).
1
2
2
zz vt
=+ −
t
0
z
0
Since the ground is flat and level,
z
and
z
0
are equal to each other. The time it takes the
golf ball to return the ground is equal to
2
z
v
t
=
0
(5.37)
g
The horizontal distance the ball will travel before it hits the ground can be computed
from Equation (5.8a) using the expression for time from Equation (5.37) and the fact
that
v
x
0
= 30 and
v
z
0
= 20.
2
vv
Δ= − =
xxx vt
=
xz
00
=
122.4
m
0
x
0
g
3.
The initial velocity of the arrow can be calculated from Equation (29).
eFx
m
s
v
=
=
53.5
mkM
+
