Respuesta :
Answer:
Explanation:
Let's break this down into parts:
Given:
Initial velocity (V₀) = 20 m/s
Angle of projection (θ) = 30 degrees
A. To find the horizontal and vertical components of the initial velocity:
The horizontal component of velocity (V₀x) = V₀ * cos(θ)
The vertical component of velocity (V₀y) = V₀ * sin(θ)
Let's calculate these values:
V₀x = 20 m/s * cos(30°)
V₀x = 20 m/s * √3/2
V₀x = 10√3 m/s ≈ 17.32 m/s (rounded to two decimal places)
V₀y = 20 m/s * sin(30°)
V₀y = 20 m/s * 1/2
V₀y = 10 m/s
B. To find the time of flight:
The time of flight can be calculated using the vertical component of velocity and gravity.
The formula for the time of flight (T) when an object is projected upward and then comes back to the same level is:
�
=
2
∗
�
0
�
�
T=
g
2∗V
0y
, where
�
g is the acceleration due to gravity (approximately 9.81 m/s²).
�
=
2
∗
10
�
/
�
9.81
�
/
�
2
T=
9.81m/s
2
2∗10m/s
�
≈
2.04
�
T≈2.04s
C. To find the range:
The range of the projectile can be calculated using the formula:
�
=
�
0
�
∗
�
R=V
0x
∗T
�
=
17.32
�
/
�
∗
2.04
�
R=17.32m/s∗2.04s
�
≈
35.36
�
R≈35.36m
D. To find the horizontal displacement at
�
=
1.5
t=1.5 s:
The horizontal displacement (
�
x) at any time
�
t can be found using the formula:
�
=
�
0
�
∗
�
x=V
0x
∗t
�
=
17.32
�
/
�
∗
1.5
�
x=17.32m/s∗1.5s
�
≈
25.98
�
x≈25.98m
So, the answers would be:
A. Horizontal component of initial velocity:
10
3
10
3
m/s (approximately 17.32 m/s), Vertical component of initial velocity: 10 m/s
B. Time of flight: Approximately 2.04 seconds
C. Range: Approximately 35.36 meters
D. Horizontal displacement at
�
=
1.5
t=1.5 sLet's break this down into parts:
Given:
Initial velocity (V₀) = 20 m/s
Angle of projection (θ) = 30 degrees
A. To find the horizontal and vertical components of the initial velocity:
The horizontal component of velocity (V₀x) = V₀ * cos(θ)
The vertical component of velocity (V₀y) = V₀ * sin(θ)
Let's calculate these values:
V₀x = 20 m/s * cos(30°)
V₀x = 20 m/s * √3/2
V₀x = 10√3 m/s ≈ 17.32 m/s (rounded to two decimal places)
V₀y = 20 m/s * sin(30°)
V₀y = 20 m/s * 1/2
V₀y = 10 m/s
B. To find the time of flight:
The time of flight can be calculated using the vertical component of velocity and gravity.
The formula for the time of flight (T) when an object is projected upward and then comes back to the same level is:
�
=
2
∗
�
0
�
�
T=
g
2∗V
0y
, where
�
g is the acceleration due to gravity (approximately 9.81 m/s²).
�
=
2
∗
10
�
/
�
9.81
�
/
�
2
T=
9.81m/s
2
2∗10m/s
�
≈
2.04
�
T≈2.04s
C. To find the range:
The range of the projectile can be calculated using the formula:
�
=
�
0
�
∗
�
R=V
0x
∗T
�
=
17.32
�
/
�
∗
2.04
�
R=17.32m/s∗2.04s
�
≈
35.36
�
R≈35.36m
D. To find the horizontal displacement at
�
=
1.5
t=1.5 s:
The horizontal displacement (
�
x) at any time
�
t can be found using the formula:
�
=
�
0
�
∗
�
x=V
0x
∗t
�
=
17.32
�
/
�
∗
1.5
�
x=17.32m/s∗1.5s
�
≈
25.98
�
x≈25.98m
So, the answers would be:
A. Horizontal component of initial velocity:
10
3
10
3
m/s (approximately 17.32 m/s), Vertical component of initial velocity: 10 m/s
B. Time of flight: Approximately 2.04 seconds
C. Range: Approximately 35.36 meters
D. Horizontal displacement at
�
=
1.5
t=1.5 seconds: Approximately 25.98 meters
econds: Approximately 25.98 meters
