[kom
22 A projectile is launched with an initial velocity of 30 m/s at an angle of
60° above the horizontal. Calculate the magnitude and direction of its
velocity 5.0 s after launch.
Answers
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.