A projectile, fired with unknown initial velocity, lands 20 s later on the side of a hill, 3000
m away horizontally and 450 m vertically above its starting point. (a) What is the vertical
component of its initial velocity? (b) What is the horizontal component of its initial
velocity
Answers
Answer:
Explanation: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