9. A plane, diving with constant speed at an angle of 53.0° with the vertical, releases a projectile at an altitude of 730 m. The projec
hits the ground 5.00 s after its release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during
flight? What are the (c) horizontal and (d) vertical components of its velocity just before striking the ground?
Answers
Explanation:
We adopt the positive direction choices used in the textbook so that equations such as Eq. are directly applicable. The coordinate origin is at ground level directly below the release point. We write θ
0
=−37.0
o
for the angle measured from +x, since the angle φ
0
=53.0
o
given in the problem is measured from the –y direction. The initial setup of the problem is shown in the figure below.
(a) The initial speed of the projectile is the plane’s speed at the moment of release. Given
that y
0
=730m and y=0 at t=5.00s, we use Eq. 4-22 to find v
0
:
y−y
0
=(v
0
sinθ
0
)t−
2
1
gt
2
⇒0.730m=v
0
sin(−37.0
o
)(5.00s)−
2
1
(9.80m/s
2
)(5.00s)
2
which yields v
0
=202m/s.
(b) The horizontal distance traveled is
R=v
x
t=(v
0
cosθ
0
)t=[(202m/s)cos(−37.0
o
)](5.00s)=806m
(c) The x component of the velocity (just before impact) is
v
x
=v
0
cosθ
0
=(202m/s)cos(−37.0
o
)=161m/s.
(d) The y component of the velocity (just before impact) is
v
y
=v
0
sinθ
0
−gt=(202m/s)sin(−37.0
o
)−(9.80m/s
2
)(5.00)=−171m/s.
Note that in this projectile problem we analyzed the kinematics in the vertical and
horizontal directions separately since they do not affect each other. The x-component of
the velocity, v
x
=v
0
cosθ
0
, remains unchanged throughout since there’s no horizontal
acceleration.