Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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)