A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-60). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance. FIGURE 3-60 Problem 88. 88. Choose the origin to be the point from which the projectile is launched, and choose upward as the positive y direction. The y displacement of the projectile is 135 m, and the horizontal range of the projectile is 195 m. The acceleration in the y direction is a y = − g , and the time of flight is 6.6 s. The horizontal velocity is found from the horizontal motion at constant velocity. Δ x = v x t → v x = Δ x t = 195 m 6.6 s = 29.55 m/s Calculate the initial v velocity from the given data and Eq. 2-12b. y = y 0 + v y 0 t + 1 2 a y t 2 → 135 m = v y 0 ( 6.6 s ) + 1 2 ( − 9.80 m/s) 2 = 60 m/s → v y 0 = 52.79 m/s Thus the initial velocity and direction of the projectile are as follows. v 0 = v x 2 + v y 0 2 = ( 29.55 m/s) 2 + (52 .79 m/s) 2 = 60m/s θ = tan − 1 v y 0 v x = tan − 1 52.79 m/s 29.55 m/s = 61 °
A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-60). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance. FIGURE 3-60 Problem 88. 88. Choose the origin to be the point from which the projectile is launched, and choose upward as the positive y direction. The y displacement of the projectile is 135 m, and the horizontal range of the projectile is 195 m. The acceleration in the y direction is a y = − g , and the time of flight is 6.6 s. The horizontal velocity is found from the horizontal motion at constant velocity. Δ x = v x t → v x = Δ x t = 195 m 6.6 s = 29.55 m/s Calculate the initial v velocity from the given data and Eq. 2-12b. y = y 0 + v y 0 t + 1 2 a y t 2 → 135 m = v y 0 ( 6.6 s ) + 1 2 ( − 9.80 m/s) 2 = 60 m/s → v y 0 = 52.79 m/s Thus the initial velocity and direction of the projectile are as follows. v 0 = v x 2 + v y 0 2 = ( 29.55 m/s) 2 + (52 .79 m/s) 2 = 60m/s θ = tan − 1 v y 0 v x = tan − 1 52.79 m/s 29.55 m/s = 61 °
A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-60). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.
FIGURE 3-60
Problem 88.
88. Choose the origin to be the point from which the projectile is launched, and choose upward as the positive y direction. The y displacement of the projectile is 135 m, and the horizontal range of the projectile is 195 m. The acceleration in the y direction is ay = −g, and the time of flight is 6.6 s.
The horizontal velocity is found from the horizontal motion at constant velocity.
Δ
x
=
v
x
t
→
v
x
=
Δ
x
t
=
195
m
6.6
s
=
29.55
m/s
Calculate the initial v velocity from the given data and Eq. 2-12b.
y
=
y
0
+
v
y
0
t
+
1
2
a
y
t
2
→
135
m
=
v
y
0
(
6.6
s
)
+
1
2
(
−
9.80
m/s)
2
= 60 m/s
→
v
y
0
=
52.79
m/s
Thus the initial velocity and direction of the projectile are as follows.
v
0
=
v
x
2
+
v
y
0
2
=
(
29.55
m/s)
2
+ (52
.79 m/s)
2
=
60m/s
θ
=
tan
−
1
v
y
0
v
x
=
tan
−
1
52.79
m/s
29.55
m/s
=
61
°
A shell is fired from a cliff that is 36 m above a horizontal plane. The muzzle speed of the shell is 80.0 m/s and it is fired at an elevation of 25° above the horizontal.
(a) Determine the horizontal range of the shell.(b) Determine the velocity of the shell as it strikes the ground.
3.00 m,
37P The three vectors in Fig. 3-31 have magnitudes a =
b = 4.00 m, and c =
10.0 m. What are (a) the x component and
(b) the y component of ā; (c) the
x component and (d) the y com-
ponent of b; and (e) the x com-
ponent and (f) the y component
of ĉ? If & = pā + qb, what are
the values of (g) p and (h) q? ilw
30°
a
Fig. 3-31 Problem 37.
A man stands on the roof of a 15.0 m tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 35° above the horizontal. You can ignore air resistance. Calculate:
a)the maximum height above the roof reached by the rock.
b)the magnitude of the velocity of the rock just before it strikes the ground
c)the horizontal range from the base of the building to the point where the rock strikes the ground.
Chapter 3 Solutions
Physics for Scientists and Engineers with Modern Physics
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