Verified answer

First, decompose the initial velocity in a horizontal and vertical component.

Horizontal component:

Vx = 31.2 · cos(40) = 23.9 m/s

Initial vertical component:

Viy = 31.2 · sin(40) = 20.1 m/s

Now you can find the time it takes to hit the ground.

d = V·t + a·t²/2

where

d = distance = -720 (the package must fall that distance down)

V = initial vertical velocity = Viy = 20.1 m/s

t = time = ?

a = acceleration by gravity = -9.8 m/s²

so

-720 = 20.1·t + (-9.8)·t²/2

4.9t² - 20.1t - 720 = 0

This is a quadratic equation that you can solve for t, using the discriminant.

√D = √((-20.1)² - 4 · 4.9 · (-720)) = 120.48

t1 = (20.1 + 120.48) / (2 · 4.9) = 14.34 s

t2 = (20.1 - 120.48) / (2 · 4.9) = -10.24 s

A negative time doesn't make sense in this case, so there is only one valid solution:

t = 14.34 s

That's the time the package needs to hit the ground.

The horizontal distance the package traveled, is:

r = V · t

where

r = horizontal distance = ?

V = initial horiziontal velocity = Viy = 20.1 m/s

t = time = 14.34 s

so

r = 20.1 · 14.34

r = 288.2 m <- - - - - - - - - - - - - - - - answer a

Now find the vertical component of the final velocity.

Vfy = Viy + a · t

where

Vfy = final vertical velocity = ?

Viy = initial vertical velocity = 20.1 m/s

a = acceleration by gravity = -9.8 m/s²

t = time = 14.34 s

so

Vfy = 20.1 + (-9.8) · 14.34

Vfy = -120.4 m/s

To find the angle of impact, use this formula:

tan(θ) = | Vfy / Vx |

where

θ = angle of impact = ?

Vfy = final vertical velocity = -120.4 m/s

Vx = horizontal velocity = 23.9 m/s

so

tan(θ) = | -120 / 23.9 |

θ = arctan(5)

θ = 78.8° <- - - - - - - - - - - - - - - - answer b