Question

A plane takes off from an airport at an angle of 14.5°. If the plane is traveling at a velocity of 220 feet per second, how high off of the ground is the plane at 10 seconds?

Answer 1

The distance of the plane travelling at 220 feet per second off the ground is 151.71feet

How to calculate the maximum height of an object

Maximum height of the plane is expressed as:

$M=\frac{u^2sin^2\theta}{2g}$

u  is the velocity = 220ft/s

θ is the angle of elevation = 14.5°.

g is the acceleration due to gravity

Substitute

$M=\frac{u^2sin^2\theta}{2g}\\M=\frac{220^2sin^2(14.5)}{2(10)}\\M=151.71feet$

Hence the distance of the plane off the ground is 151.71feet

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