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Ganezh [65]
3 years ago
9

A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

Physics
1 answer:
7nadin3 [17]3 years ago
3 0

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

v_y=v\ sin(25)

v_y=22\ m/s\times\ sin(25)

v_y=9.29\ m/s

Hence, the vertical component of the velocity is 9.29 m/s

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Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

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A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
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3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
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Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

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3 years ago
In Paragraph 3 of this passage, what clues help you know the meaning of the word
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