Question

An airplane flies horizontally with a constant speed of 172.0 m/s at an altitude of 1390 m. A package is dropped out of the airplane. Ignore air resistance. The magnitude of the gravitational acceleration is 9.8 m/s2. Choose the RIGHT as positive x-direction. Choose UPWARD as positive y-direction Keep 2 decimal places in all answers
(a) What is the vertical component of the velocity (in m/s) just before the package hits the ground? Pay attention to the direction (the sign).
(b) What is the magnitude of the velocity (in m/s) (including both the horizontal and vertical components) of the package just before it hits the ground?

Answer 1

Answer:

(a) - 165.032 m/s

(b) 238.37 m/s

Explanation:

initial horizontal velocity, ux = 172 m/s

height, h = 1390 m

g = 9.8 m/s^2

Let it strikes the ground after time t.

Use second equation of motion in vertical direction

$s=ut+\frac{1}{2}at^{2}$

-1390 = 0 - 0.5 x 9.8 x t^2

t = 16.84 second

(a) Let vy be the vertical component of velocity as it strikes the ground

Use first equation of motion in vertical direction

vy = uy - gt

vy = 0 - 9.8 x 16.84

vy = - 165.032 m/s

Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.

(b)

The horizontal component of velocity remains constant throughout the motion.

vx = 172 m/s

vy = - 165.032 m/s

The resultant velocity is v.

$v=\sqrt{172^{2}+165.032^{2}}$

v = 238.37 m/s

Thus, teh velocity with which it hits the ground is 238.37 m/s.