- #1

EHogeberg

- 9

- 0

## Homework Statement

A projectile is fired with an initial speed of 16.0 m/s at an angle of 45.0^\circ above the horizontal. The object hits the ground 9.00 s later.

What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Vo=16m/s

a=-g

θ=45

Tf=9 seconds

magnitude of velocity an instant before it hits the ground: 77.7 m/s

max height: 301.53 m/s

## Homework Equations

-gsin(θ)

Vxf=Vxi+a(tf-ti)

Xf=Xi+Vxi(tf-ti) + 1/2 A(tf-ti)^2

Vxf^2=Vxi^2+2a(Xf-Xi)

Vyf=Vyi+a(tf-ti)

Yf=yi+Vyi(tf-ti) + 1/2 A(tf-ti)^2

Vyf^2=Vyi^2+2a(Yf-Yi)

## The Attempt at a Solution

I got 45 degrees and I got that from finding the xcomponents and ycomponents