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The Velocity of the Projectile as It Strikes the Ground
The velocity of the projectile as it hits the ground v can be described in terms of its components, as shown in
g
figure 3.25. The x-component of the velocity at the ground, found from equation 3.40, is
v = v = v (3.50)
xg x 0x
The y-component of the velocity at the ground, found from equation 3.41 with t = t
t
is
v = v - gt
yg 0y t
= v - g(2v ) (3.51)
0y 0y
g
v = -v (3.52)
yg 0y
Figure 3.25 The velocity of the
projectile at the ground.
The y-component of the velocity of the projectile at the ground is equal to the negative of the y-component of the
original velocity. The minus sign just indicates that the projectile is coming down. But this is exactly what we
expected from the study of one-dimensional motion. The magnitude of the actual velocity at the ground, found
from its two components, is
2 2
vg = vxg + vyg (3.53)
( ) ( )
and using equations 3.50 and 3.52, becomes
2
2
vg = v0x + (3.54)
( ) (-v0 y = v0
)
The speed of the projectile as it strikes the ground is equal to the original speed of the projectile. The direction that
the velocity vector makes with the ground is
¸ = tan-1 v = tan-1 -v = -¸
yg 0y
v v
xg 0x
The angle that the velocity vector makes as it hits the ground is the negative of the original angle. That is, if the
projectile was fired at an original angle of 300 above the positive x-axis, it will make an angle of 300 below the
positive x-axis when it hits the ground.
The Location and Velocity of the Projectile at Any Time t
We find the position and velocity of the projectile at any time t by substituting that value of t into equations 3.38,
3.39, 3.40, and 3.41. Let us look at some examples of projectile motion.
Example 3.10
Projectile motion in two dimensions. A ball is thrown with an initial velocity of 30.0 m/s at an angle of 60.00 above
the horizontal, as shown in figure 3.26. Find (a) the maximum height of the ball, (b) the time to rise to the top of
the trajectory, (c) the total time the ball is in the air, (d) the range of the ball, (e) the velocity of the ball as it
strikes the ground, and (f) the position and velocity of the ball at t = 4 s.
Solution
The x-component of the initial velocity is
v = v cos ¸ = (30.0 m/s) cos 600 = 15.0 m/s
0x 0
The y-component of the initial velocity is
3-28 Mechanics
v = v sin ¸ = (30.0 m/s) sin 60.00 = 26.0 m/s
0y 0
a. The maximum height of the ball, found from
equation 3.49, is
y = v = (26.0 m/s)2 = 34.5 m
max 0y2
2g 2(9.80 m/s2)
b. To find the time to rise to the top of the
trajectory, we use equation 3.43,
t = v = 26.0 m/s = 2.65 s
r 0y
g 9.80 m/s2
Figure 3.26 Trajectory of a thrown ball.
c. To find the total time that the ball is in the air, we use equation 3.45,
t = 2t = 2(2.65 s) = 5.30 s
t r
d. The range of the ball, found from equation 3.47, is
R = v sin 2¸ = (30.0 m/s)2 sin 1200 = 79.5 m
02
g 9.80 m/s2
As a check, we can use equation 3.46 to get
R = x = v t = (15.0 m/s)(5.30 s) = 79.5 m
max 0x t
e. To find the magnitude of the velocity of the ball at the ground, we use equation 3.53,
2 2
vg = vxg + vyg
( ) ( )
where
v = v = 15.0 m/s
xg 0x
and
v = v - gt = 26.0 m/s - (9.80 m/s2)(5.30 s)
yg 0y t
= 26.0 m/s - 51.9 m/s = -25.9 m/s
Hence,
22
vg = 15.0 m/s +
() (-25.9 m/s
)
= 29.9 m/s E" 30.0 m/s because of round off errors
The direction that the velocity vector makes with the ground is
¸ = tan-1v = tan-1 -25.9 m/s = -59.90 E" 60.00
yg
v 15.0 m/s
xg
f. To find the position and velocity of the ball at t = 4 s we use the kinematic equations 3.38 through 3.41.
1. x = v t = (15.0 m/s)(4 s) = 60.0 m
0x
2. y = v t - 1 gt2
0y
2
= (26.0 m/s)(4 s) - 1 (9.80 m/s2)(4 s)2
2
= 25.6 m
The ball is 60.0 m down range and is 25.6 m high.
Chapter 3 Kinematics - The Study of Motion 3-29
The components of the velocity at 4 s are
3. v = v = 15.0 m/s
x 0x
4. v = v - gt
y 0y
= 26.0 m/s - (9.80 m/s2)(4 s)
= -13.2 m/s
At the end of 4 s the x-component of the velocity is 15.0 m/s and the y-component is -13.2 m/s. To determine the
magnitude of the velocity vector at 4 s we have
2
2
v = vx + vy
( ) ( )
22
v = 15.0 m/s +
() (-13.2 m/s
)
= 20.0 m/s
The direction of the velocity vector at 4 s is
¸ = tan-1v = tan-1 -13.2 m/s = -41.30
yg
v 15.0 m/s
xg
The velocity vector makes an angle of 41.30 below the horizontal at 4 s.
To go to this interactive example click on this sentence.
Example 3.11
A projectile is fired horizontally from the roof of a building. A projectile is fired horizontally from the roof of a
building 30.0 m high at an initial velocity of 20.0 m/s, as shown in figure 3.27. Find (a) the total time the projectile
is in the air, (b) where the projectile will hit the ground, and (c) the velocity of the projectile as it hits the ground.
Solution
The x- and y-components of the velocity are
v = v = 20.0 m/s
0x 0
v = 0
0y
a. To find the total time that the projectile is in the air, we use
equation 3.39,
y = v t - 1 gt2
0y
2
However, the initial conditions are that v = 0. Therefore,
0y
y = - 1 gt2
2
Solving for t,
2y
t = -
g
Figure 3.27 Trajectory of projectile thrown horizontally.
However, when t = t , y = -30.0 m. Hence,
t
2
)
2y (-30.0 m
tt = - = -
g 9.80 m/s2
= 2.47 s
3-30 Mechanics
b. To find where the projectile hits the ground, we use equation 3.38,
x = v t
0x
Now the projectile hits the ground when t = t , therefore,
t
x = v t = (20.0 m/s)(2.47 s) = 49.4 m
0x t
The projectile hits the ground at the location y = -30.0 m and x = 49.4 m.
c. To find the velocity of the projectile at the ground we use equations 3.50, 3.51, and 3.53:
v = v = v = 20.0 m/s
xg 0x 0
v = v - gt = 0 - (9.80 m/s2)(2.47 s) = -24.2 m/s
yg 0y t
2 2
vg = vxg + vyg
( ) ( )
22
v = 20.0 m/s +
() (-24.2 m/s
)
= 31.4 m/s
The direction that the velocity vector makes with the ground is
¸ = tan-1v = tan-1 -24.2 m/s = -50.40
yg
v 20.0 m/s
xg
The velocity vector makes an angle of 50.40 below the horizontal when the projectile hits the ground.
To go to this interactive example click on this sentence.
Example 3.12
A projectile is fired at an angle from the roof of a building. A projectile is fired at an initial velocity of 35.0 m/s at
an angle of 30.00 above the horizontal from the roof of a building 30.0 m
high, as shown in figure 3.28. Find (a) the maximum height of the
projectile, (b) the time to rise to the top of the trajectory, (c) the total
time that the projectile is in the air, (d) the velocity of the projectile at
the ground, and (e) the range of the projectile.
Solution
The x- and y-components of the original velocity are
v = v cos ¸ = (35.0 m/s) cos 300 = 30.3 m/s
0x 0
v = v sin ¸ = (35.0 m/s) sin 300 = 17.5 m/s
0y 0
a. To find the maximum height we use equation 3.49:
Figure 3.28 Trajectory of a projectile fired
from the roof of a building.
y = v = (17.5 m/s)2
max 0y2
2g 2(9.80 m/s2)
= 15.6 m
above the building. Since the building is 30 m high, the maximum height with respect to the ground is 45.6 m.
b. To find the time to rise to the top of the trajectory we use equation 3.43:
Chapter 3 Kinematics - The Study of Motion 3-31
t = v = 17.5 m/s = 1.79 s
r 0y
g 9.80 m/s2
c. To find the total time the projectile is in the air we use equation 3.39:
y = v t - 1 gt2
0y
2
When t = t , y = -30.0 m. Therefore,
t
-30.0 m = (17.5 m/s)t - 1 (9.80 m/s2)t
t t2
2
Rearranging the equation, we get
4.90 t - 17.5 t - 30.0 = 0
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