Answer:
7.09683 m
1.20285 s
2.4057 s
11.8 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)
From equation of motion we have
The maximum height above the ground that the ball reaches is 7.09683 m
Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s
The velocity just before it hits the ground is 11.8 m/s
Answer:
The force required to move the quarterback with linebacker is <u>1215 N</u>
Explanation:
Using Newton's second law, it is established that F = Ma
Where F is net force acting on the system, a is the acceleration and M is mass of the two object 
Now consider both 
as a system, so net force acting on the system is 
Substitute the given values in the above formula,
Force = 1215 N
<u>1215 N </u>is the force required to move the quarterback with linebacker.
Answer:
Acceleration is 0.25m/s^2
Explanation:
Given the following :
Speed = 0.5m/s
Radius(r) of circle = 1m
Acceleration round a circular path is given as :
a = v^2 / r
Where
a = acceleration of the body
v = speed / velocity
r = radius
Therefore,
a = v^2 / r
a = (0.5)^2 / 1
a = 0.25m/s^2
Answer:Bruce is knocked backwards at
14
m
s
.
Explanation:
This is a problem of momentum (
→
p
) conservation, where
→
p
=
m
→
v
and because momentum is always conserved, in a collision:
→
p
f
=
→
p
i
We are given that
m
1
=
45
k
g
,
v
1
=
2
m
s
,
m
2
=
90
k
g
, and
v
2
=
7
m
s
The momentum of Bruce (
m
1
) before the collision is given by
→
p
1
=
m
1
v
1
→
p
1
=
(
45
k
g
)
(
2
m
s
)
→
p
1
=
90
k
g
m
s
Similarly, the momentum of Biff (
m
2
) before the collision is given by
→
p
2
=
(
90
k
g
)
(
7
m
s
)
=
630
k
g
m
s
The total linear momentum before the collision is the sum of the momentums of each of the football players.
→
P
=
→
p
t
o
t
=
∑
→
p
→
P
i
=
→
p
1
+
→
p
2
→
P
i
=
90
k
g
m
s
+
630
k
g
m
s
=
720
k
g
m
s
Because momentum is conserved, we know that given a momentum of
720
k
g
m
s
before the collision, the momentum after the collision will also be
720
k
g
m
s
. We are given the final velocity of Biff (
v
2
=
1
m
s
) and asked to find the final velocity of Bruce.
→
P
f
=
→
p
1
f
+
→
p
2
f
→
P
f
=
m
1
v
1
f
+
m
2
v
2
f
Solve for
v
1
:
v
1
f
=
→
P
f
−
m
2
v
2
f
m
1
Using our known values:
v
1
f
=
720
k
g
m
s
−
(
90
k
g
)
(
1
m
s
)
45
k
g
v
1
f
=
14
m
s
∴
Bruce is knocked backwards at
14
m
s
.
Explanation:
