A body is projected at an angle of 45° with the horizontal with a velocity of 60sqrt(2) msThen the angle made by its velocity with the horizontal after 6s is
I need in degrees
If u can do it u are great
If u cant waste
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A body is projected at an angle of 45° with the horizontal with a velocity of 60sqrt(2) msThen the angle made by its velocity with the horizontal after 6s is
I need in degrees
If u can do it u are great
If u cant waste
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Answer:
Certainly, I can help with that!
The horizontal and vertical components of the initial velocity (\( u \)) can be found using trigonometric functions. Given that the angle of projection (\( \theta \)) is 45 degrees, the horizontal component (\( u_x \)) is \( u \cdot \cos(\theta) \) and the vertical component (\( u_y \)) is \( u \cdot \sin(\theta) \).
\[ u_x = 60\sqrt{2} \cdot \cos(45^\circ) \]
\[ u_y = 60\sqrt{2} \cdot \sin(45^\circ) \]
After 6 seconds, the horizontal component (\( v_x \)) remains constant, but the vertical component (\( v_y \)) changes due to the acceleration due to gravity.
\[ v_x = u_x \]
\[ v_y = u_y - gt \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( t \) is the time of flight (6 seconds in this case).
Now, you can find the angle (\( \phi \)) the velocity vector makes with the horizontal using the arctangent function:
\[ \phi = \tan^{-1}\left(\frac{v_y}{v_x}\right) \]
Finally, convert the angle from radians to degrees.
Let me calculate that for you.
\[ u_x = 60\sqrt{2} \cdot \cos(45^\circ) \approx 60 \, \text{m/s} \]
\[ u_y = 60\sqrt{2} \cdot \sin(45^\circ) \approx 60 \, \text{m/s} \]
\[ v_x = u_x = 60 \, \text{m/s} \]
\[ v_y = u_y - gt = 60 \, \text{m/s} - (9.8 \, \text{m/s}^2 \cdot 6 \, \text{s}) \approx -14.8 \, \text{m/s} \]
\[ \phi = \tan^{-1}\left(\frac{v_y}{v_x}\right) \approx \tan^{-1}\left(\frac{-14.8}{60}\right) \approx -13.8^\circ \]
So, the angle made by the velocity with the horizontal after 6 seconds is approximately \( -13.8^\circ \) (negative because it's below the horizontal). If you have any further questions, feel free to ask!
Answer:
Explanation:
When a projectile is launched at an angle with the horizontal, we can use the kinematic equations to find the components of its velocity at any given time. The horizontal and vertical components of the initial velocity (
�
u) are given by:
�
�
=
�
cos
�
u
x
=ucosθ
�
�
=
�
sin
�
u
y
=usinθ
where:
�
u is the initial velocity,
�
θ is the angle of projection.
In this case, the initial velocity (
�
u) is given as
60
2
�
/
�
60
2
m/s and the angle of projection (
�
θ) is
4
5
∘
45
∘
.
�
�
=
60
2
cos
4
5
∘
u
x
=60
2
cos45
∘
�
�
=
60
2
sin
4
5
∘
u
y
=60
2
sin45
∘
After
�
t seconds, the horizontal and vertical components of velocity (
�
�
v
x
and
�
�
v
y
) are given by:
�
�
=
�
�
v
x
=u
x
�
�
=
�
�
−
�
�
v
y
=u
y
−gt
where:
�
g is the acceleration due to gravity (
9.8
�
/
�
2
9.8 m/s
2
),
�
t is the time of flight.
Now, let's calculate
�
�
v
x
and
�
�
v
y
after
6
�
6 s:
�
�
=
60
2
cos
4
5
∘
v
x
=60
2
cos45
∘
�
�
=
60
2
sin
4
5
∘
−
(
9.8
�
/
�
2
)
⋅
(
6
�
)
v
y
=60
2
sin45
∘
−(9.8 m/s
2
)⋅(6 s)
Next, we can find the angle
�
α that the velocity makes with the horizontal using the arctangent function:
tan
�
=
�
�
�
�
tanα=
v
x
v
y
Finally, the angle
�
α is the angle made by the velocity with the horizontal after
6
�
6 s.
Let's calculate this:
�
=
arctan
(
�
�
�
�
)
α=arctan(
v
x
v
y
)
�
=
arctan
(
60
2
sin
4
5
∘
−
(
9.8
�
/
�
2
)
⋅
(
6
�
)
60
2
cos
4
5
∘
)
α=arctan(
60
2
cos45
∘
60
2
sin45
∘
−(9.8 m/s
2
)⋅(6 s)
)
Now, compute the numerical value of
�
α, and it will give you the angle in degrees.