In a football game,a player wants to hit a footbal are in the plane of field. Football's initial velocity v football lands on ground?
Answers
Answer:
the answer to your question is 0.5
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Answer:
We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the initial position for the football as it begins projectile motion in the sense of §4-5), and we let θ
0
be the angle of its initial velocity measured from the +x axis.
(a) x=46m and y=–1.5m are the coordinates for the landing point; it lands at time t=4.5s. Since x=v
0x
t,v 0x= tx= 4.5s
46m
=10.2m/s.
Since y=v
0y
t−
2
1
gt
2
,
v
0y
=
t
y+
2
1
gt
2
=
4.5s
(−1.5m)+
2
1
(9.8m/s
2
)(4.5s)
2
=21.7m/s.
The magnitude of the initial velocity is
v
0
=
v
0x
2
+v
0y
2
=
(10.2m/s)
2
+(21.7m/s)
2
=24m/s.
(b) The initial angle satisfies tanθ
0
=v
0y
/v
0x
. Thus,
θ
0
=tan
−1
[(21.7m/s)/(10.2m/s)]=65
o
.
Explanation:
We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the initial position for the football as it begins projectile motion in the sense of §4-5), and we let θ
0
be the angle of its initial velocity measured from the +x axis.
(a) x=46m and y=–1.5m are the coordinates for the landing point; it lands at time t=4.5s. Since x=v
0x
t,
v
0x
=
t
x
=
4.5s
46m
=10.2m/s.
Since y=v
0y
t−
2
1
gt
2
,
v
0y
=
t
y+
2
1
gt
2
=
4.5s
(−1.5m)+
2
1
(9.8m/s
2
)(4.5s)
2
=21.7m/s.
The magnitude of the initial velocity is
v
0
=
v
0x
2
+v
0y
2
=
(10.2m/s)
2
+(21.7m/s)
2
=24m/s.
(b) The initial angle satisfies tanθ
0
=v
0y
/v
0x
. Thus,
θ
0
=tan
−1
[(21.7m/s)/(10.2m/s)]=65
o
.
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