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
°
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Chapter 3 Solutions
Physics for Scientists and Engineers with Modern Physics
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