A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?
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Answer:
A projectile, fired with unknown initial velocity, lands 18.4 s later on the side of a hill, 2740...
Question:
A projectile, fired with unknown initial velocity, lands 18.4 s later on the side of a hill, 2740 m away horizontally and 469 m vertically above its starting point. (Ignore any effects due to air resistance.)
(a) What is the vertical component of its initial velocity? m/s
(b) What is the horizontal component of its initial velocity? m/s
(c) What was its maximum height above its launch point? m
(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical? v = m/s ? =
Projectile Motion:
It is a two-dimensional motion of a particle that is thrown with some initial velocity and under the influence of gravity follows a curved parabolic path.
Let the particle is thrown at a velocity of 'u' m/s and at an angle of
θ
with the horizontal. The acceleration due gravity is acting vertically downwards and equals to
9.8
m
/
s
2
.
Using the equations of motion in two perpendicular directions namely x and y axis to get the simpler and direct formulas.
Time of Flight: The time interval when the particle was in the air.
The initial velocity of particle in y-direction will be:
u
sin
θ
.
Applying displacement equation to get the time of flight:
Since, the particle returns to y = 0 after full journey. So the displacement of the particle in y-direction is zero:
Δ
s
=
0
Therefore,
Δ
s
=
u
y
t
+
0.5
a
t
2
0
=
(
u
sin
θ
)
T
−
0.5
×
g
T
2
T
=
2
u
sin
θ
g
Range: Horizontal distance covered by the particle is its range.
The particle has initial horizontal velocity as
u
cos
θ
and it is constant throughout the journey as the particle does not have any acceleration in horizontal direction.
Applying displacement equation and putting the time we got above to get the range:
Δ
s
=
u
x
t
+
0.5
a
t
2
R
=
(
u
cos
θ
)
T
+
0
R
=
(
u
cos
θ
)
×
2
u
sin
θ
g
R
=
u
2
sin
2
θ
g
Height: The maximum height achieved by the particle throughout the journey is called as the height of projectile.
Solve for height along the y-direction. So, at the maximum height velocity of particle along y-axis is zero.
Applying equation of motion:
Concept:
When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration directed towards the center of the earth, that curve path of the particle is called a projectile, and the motion is called projectile motion.
Given:
The time of flight, t = 20 sec
Horizontal Range, R = 3000 m
Maximum Height, H = 450 m
Find:
The vertical and horizontal components of the initial velocity that is given to the projectile.
Solution:
In a projectile motion, moving with a velocity, u, u cosθ is the horizontal component and u sinθ is the vertical component of the velocity.
The time of flight can be defined as,
t =2×u sinθ/g
From this, u sinθ can be calculated,
The acceleration due to gravity is taken as g = 10 m/s²
u sinθ = tg/2 = 20 × 10/2 = 100 m/s
The horizontal range can be defined as,
R = u²sin2θ /g
R = u cos θ × 2u sinθ/g
As t =2×u sinθ/g = 20 sec and R = 3000m,
R = u cosθ × t
u cosθ = R/t = 3000/20 = 150 m/s
Hence, the vertical component of the initial velocity is 100 m/s and the horizontal component of the initial velocity is 150 m/s.
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