So I want to find the speed an object hits the ground.

I have the initial speed 45ms, the angle 50degrees and its time in flight 6.7s.

So the final speed before impact should be:
(45ms)sin(50)-9.81(6.7)=-31.25...ms

Does this seem correct. In my head I was expecting that the acceleration due to gravity on the return to ground would increase the speed beyond its initial speed (considering im not factoring air resistance). Does this seem correct?

It is a parabola with the vertex at the midpoint peak height. If the landing is at the same level as the takeoff the thing will hit at the same speed as it took off.

u = horizontal speed = 45 cos 50 forever

vertical problem:
v = 45 sin 50 - g t
h = 45 sin 50 t - .5 g t^2
at ground h = 0
.5 g t^2 -45 sin 50 t = 0
t (.5 g t -45 sin 50 ) = 0
t = 0 of course, the start
and
t = 45 sin 50 /.5 g
so at crash
v = 45 sin 50 - g(45 sin 50/.5g)
v = 45 sin 50 - 2(45 sin 50)
v = - 45 sin 50
so same speed down as it started up

Aha. The landing point is 10m higher than the take off point. So this would make sense.

There is an easy way.

Doing vertical problem only:
Vi = 45 sin 50 as before
Now the vertical problem can be done for object thrown straight up and the horizontal component added in later.
Conservation of energy:
(1/2)m Vi^2 = (1/2) m v^2 + m g h
where h is the height above initial height
v^2 = Vi^2 - 2 g h

then total speed = sqrt(u^2+v^2)

The calculation for the final speed before impact that you have done is not correct. Let's go through the correct method step by step to find the final speed of the object.

First, we need to break down the initial velocity into horizontal and vertical components. The horizontal component remains constant and does not change, while the vertical component is affected by gravity.

The initial vertical velocity can be calculated using the equation:
v0y = v0 * sin(angle)
v0y = 45 m/s * sin(50 degrees) ≈ 34.64 m/s (rounded to two decimal places)

Next, we can calculate the time it takes for the object to reach the highest point of its trajectory (the time of flight divided by 2):
t_max = 6.7 s / 2 = 3.35 s

At the highest point, the vertical velocity is zero. Using this information, we can find the vertical distance traveled during this time:
S_max = v0y * t_max + (1/2) * (-9.81) * t_max^2
S_max = 34.64 m/s * 3.35 s + (1/2) * (-9.81 m/s^2) * (3.35 s)^2 ≈ 58.10 m (rounded to two decimal places)

Now, we can use the equation for uniform acceleration to find the final vertical velocity, vfy, at the time of impact:
vfy = v0y + (-9.81) * t_max
vfy = 34.64 m/s + (-9.81 m/s^2) * 3.35 s ≈ 1.29 m/s (rounded to two decimal places)

Finally, we can find the final overall velocity, vf, before impact using the Pythagorean theorem:
vf = sqrt((vfx)^2 + (vfy)^2)
vf = sqrt((v0 * cos(angle))^2 + (vfy)^2)
vf = sqrt((45 m/s * cos(50 degrees))^2 + (1.29 m/s)^2 ) ≈ 38.95 m/s (rounded to two decimal places)

Therefore, the correct final speed before impact is approximately 38.95 m/s.

Regarding your idea that the acceleration due to gravity could increase the speed beyond the initial velocity, in this case, since the object was launched upward at an angle, gravity acts against the initial velocity during the upward trajectory, causing the object to slow down. As it falls back down, gravity acts with the initial velocity, but by that time the object has lost some of its initial speed already due to the upward journey. Therefore, the final velocity is less than the initial velocity.