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

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asked Jan 4, 2021 145k views

1 Answer

Anita Jayana · Answer 1 · 2021-01-10T16:08:32+0000

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
$H = 318.5 \ m$

Step-by-step 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

$H = 318.5 \ m$

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

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