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2 years ago

An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal

Physics

1 answer:

joja [24]2 years ago

6 0

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?

Answer:

The height is $H = 318.5 \ m$

Explanation:

From the question we are told that

The speed at which the plane takes off is $u = 49 \ m/s$

The angle at which it takes off is $\theta = 30 ^o$

The time taken is $t = 13 s$

The vertical distance traveled is mathematically represented as

$H = u sin \theta t$

Substituting values

$H = (49) * sin (30) *13$

$H = 318.5 \ m$

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We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

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Therefore, the deceleration of the box is –0.08 ms¯²

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Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

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Therefore, it will take 262.5 s for the box to come to rest.

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