Question

An airplane flying at a speed of 130 mi/h (58.1 m/s) and at an altitude of 4.9 km drops a food package. Without a parachute, at what angle (degrees) below the horizontal will the package strike the ground?

Answer 1

The food package will strike the ground at 11 degrees below the horizontal.

Time for the food package to hit the ground

The time for the food package to hit the ground is calculated as follows;

h = vt + ¹/₂gt²

let the initial velocity be horizontal

4900 = 0(t) + (0.5 x 9.8)t²

4900 = 4.9t²

t² = 4900/4.9

t² = 1,000

t = √1,000

t = 31.62 s

Final speed of the food package when it hits ground

vf(y) = vo + gt

vf(y) = 0 + (31.62 x 9.8)

vf(y) = 309.88 m/s

Angle of projection

The horizontal component of the speed will be constant, while vertical component will change

$tan(\theta ) = \frac{V_y}{V_x} \\\\\theta = tan^{-1} (\frac{V_y}{V_x})\\\\\theta = tan^{-1} (\frac{309.88}{58.1} )\\\\\theta = 79^0$

Angle below the horizontal = 90 - 79 = 11⁰

Thus, the food package will strike the ground at 11 degrees below the horizontal.

Learn more about angle of projection here: brainly.com/question/10671136