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Inessa [10]
2 years ago
5

A projectile is launched with an initial speed of 21.8 m/s at an angle of 35º above the horizontal. Determine the time of flight

of the projectile.
Physics
1 answer:
katrin2010 [14]2 years ago
8 0

Answer:

2.55 seconds

Explanation:

Given that,

The initial speed of a projectile, u = 21.8 m/s

The angle of projection, \theta=35^{\circ}

We need to find the time of flight of the projectile. The formula for the time of flight is given by :

T=\dfrac{2u\sin\theta}{g}

Where

g is the acceleration due to gravity

T=\dfrac{2\times 21.8\sin(35)}{9.8}\\\\T=2.55\ s

So, the time of flight of the projectile is 2.55 seconds.

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Explanation:

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Force, F = 220 N

width, r = 1.40 m

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to find the times needed to close the door we can use the following equation

θ = ω₀t + 1/2 αt²

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t = time

in this case, the angular velocity is zero. thus,

θ = 1/2 αt²

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τ = I α

where

τ = torque

I = Inertia

we know that

τ = F r

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I = 1/3 mr²

so,

τ = I α

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