Given the setup shown in figure. A projectile is launched with initial speed Vo = 30 m/s at a given angle
a = 60° above the horizontal. At a horizontal distance d = 30 m from the launch point stands an observer,
yt
d
Suppose that at the instant the projectile is launched, the observer also launches as "interceptor vertically
upward. With what initial speed must the interceptor be launched such that it hits the projectile?
.
Answers
Answer:
ANSWER
(a) We choose horizontal x and vertical y axes such that both components of
v
0
are positive.
Positive angles are counterclockwise from +x and negative angles are clockwise from it.
In unit-vector notation, the velocity at each instant during the projectile motion is
v
=v
0
cosθ
0
i
^
+(v
0
sinθ
0
−gt)
j
^
.
putting values v
0
=30m/s and θ
0
=60°,
v
=(15
i
^
+6.4
j
^
)m/s, for t=2.0s.
The magnitude of
v
is ∣
v
∣=
(15m/s)
2
+(6.4m/s)
2
=16m/s.
(b) The direction of
v
is
θ=tan
−1
[(6.4m/s)/(15m/s)]=23
o
,
measured counterclockwise from +x.
(c) Since the angle is positive, it is above the horizontal.
(d)
v
=v
0
cosθ
0
i
^
+(v
0
sinθ
0
−gt)
j
^
.
putting values v
0
=30m/s and θ
0
=60°,
With =5.0s,
v
=(15
i
^
−23
j
^
)m/s
(e) The direction of
v
is θ=tan
−1
[(−23m/s)/(15m/s)]=−57
o
,
i.e 57° measuredclockwise from +x.
(f) Since the angle is negative, it is below the horizontal.
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