- #1

reyrey389

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## Homework Statement

An airplane with a speed of 70.6 m/s is climbing upward at an angle of 40.0 ° with respect to the horizontal. When the plane's altitude is 814 m, the pilot releases a package.

**(a)**Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.

**(b)**Relative to the ground, determine the angle of the velocity vector of the package just before impact.

## Homework Equations

70.6 * cos(40) = 54.1 velocity initial in the x direction = velocity final

70.6 * sin(40) = 45.4 velocity initial in the y direction

Displacement(Y) = -814 meters (initial to final position)

## The Attempt at a Solution

I already obtained the answer for part A. What I did was solve for t using

-814 = 45.4*t - 4.9t^2

t = 18.3 = time of flight

so for part a) the horizontal distance = 18.3 * 54.1 = 990 meters

now for part b) I have the final velocity in the x (54.1), I need the final velocity in the y, and I can use inverse tangent to find the angle once v_final(y direction) is found.

To find this value I used v_y = v_initial(y) - g*t using 45.4 for v_initial(y) and 18.3 for time

which gives v_y = -133.94 and so theta = inverse tangent ( 133.94/54.1) = 68 degrees from the ground, however this is wrong .

I'm assuming my value for the final velocity in the y direction is wrong...