A bullet was fired horizontally with 20 m/s from the top of a building 20 m high. When the bullet was 10 m above the ground, incidentally it hits a bird. Find the time taken to hit the bird and the velocity of bullet when it hits bird. Ans:(1.428 sec, 24.41 m/s)
CLASS 11 GRADE PROBLEM
CHAPTER PROJECTILE MOTION
Answers
Answer:
The initial vertical speed is
0
m
s
−
1
. Let the total time that the bullet takes to hit the bird be
T
(
s
)
. The bullet will travel vertically
10
m
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
s
=
10
m
u
=
0
m
s
−
1
v
=
Not Required
m
s
−
1
a
=
g
m
s
−
2
t
=
T
s
So we can calculate
T
using
s
=
u
t
+
1
2
a
t
2
to get:
10
=
(
0
)
(
T
)
+
1
2
(
g
)
(
T
2
)
∴
10
=
1
2
g
T
2
∴
g
T
2
=
20
∴
T
2
=
20
g
If we take
g
=
9.8
m
s
−
2
then we get:
T
2
=
20
9.8
⇒
T
=
±
1.428571
...
Hence
T
=
1.4
s
(2sf)
Explanation:
Explanation:
We will assume that the only force acting on the bullet after it is fired is that of gravity (ignore air resistance) and treat this as a simple projectile motion problem which can be solved with kinematics.
The bullet is fired with an initial horizontal velocity, but no initial vertical velocity. Therefore,
v
i
y
=
0
.
We also know that for a simple projectile, there is no horizontal acceleration, and under only the influence of gravity, the vertical acceleration is equal to
−
g
.
−
g
=
−
9.8
m
s
2
Note that
g
, the gravitational acceleration constant, is a positive value. The negative sign indicates a downward acceleration, and is used in projectile motion problems to describe a falling object.
With values for
a
y
and
v
i
y
, as well as the given
Δ
y
, we can use a kinematic equation to solve for
Δ
t
.
y
f
=
y
i
+
v
i
y
Δ
t
+
1
2
a
y
Δ
t
2
⇒
Δ
y
=
1
2
a
y
Δ
t
2
⇒
t
=
√
2
Δ
y
a
y
=
⎷
2
(
−
10
m
)
−
9.8
m
s
2
=
1.4
s
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