Projectile Motion Problem: Finding Velocity Direction at Impact

[EHogeberg]

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

 

[PeterO]

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

If it took 9 seconds to hit the ground, it must have been fired from the top of a rather tall building/high hill.

At that speed, at 45 degrees it would hit the ground [flat ground scenario] in a little over 2 seconds.

 

[NascentOxygen]

In light of what PeterO has written, I'd check that 45 degrees angle. It would have been 45 degrees after 2 secs of flight, so if it took another 7 secs, then its vertical speed would have been considerably more, making the angle steeper.