Question

A plane takes off from an airport at an angle of 13.17

If the plane is traveling at a velocity of 199 feet per second, how high off the ground is the plane at 10 seconds?

Answer 1

This problem involves trigonometry to find the height of the plane. We will need to use the sine function.

First, we need to find the vertical component of the plane's velocity. We can do this by multiplying the velocity by the sine of the angle.

sin(13.17) = 0.227 vertical velocity = 199 x 0.227 = 45.07 feet per second

Next, we can use the following kinematic equation to find the height of the plane at 10 seconds:

height = initial height + vertical velocity x time - 0.5 x acceleration x time^2

Assuming the initial height of the plane is zero (on the ground), and the acceleration due to gravity is -32.2 feet per second squared (downward), we can plug in the known values and solve for the height:

height = 0 + 45.07 x 10 - 0.5 x (-32.2) x 10^2 height = 0 + 450.7 + 1610 height = 2060.7 feet

Therefore, the plane is 2060.7 feet off the ground at 10 seconds after takeoff.