Someone double check me, basic physics problem?

Question

I just had an exam and want to know if I did this right or not. An object shoots off a building straight at 12 ft/s, and falls 260ft to the ground, what velocity does it hit at?

(1/2mVi)+(mgHi)=(1/2mVf)+(mgHf)

mass cancels
Final Height is 0

2((1/2Vi)+(gHi))=Vf

Plug in numbers and solve. I got 166.3 ft/s. Is this right or wrong?

Answer 1

the speed it gets when it falls 260 feet , the 12 ft/ sec is completely irrelevant.

Answer 2

Sorry!

The object has two components of velocity when it hits. A horizontal component of 12 ft/s and a vertical component due to the drop.

Were you asked for the velocity or speed of impact?

Velocity is not the same as speed. It is a vector quantity and you need to combine the two components as vectors.

Good luck with the rest of the exam.

Answer 3

Its a vector problem.

You can find the time it takes the object to fall vertically from s = 1/2gt^2 -> t = sqrt(2s/g). Plug this into v = gt and it hits the ground vertically at sqrt(2gs). But it ALSO has horizontal speed 12 ft/s. So you have to add these as vectors.

Answer 4

horizontal: constant 12 f/s

vertical:
initial Vo =0
dependent on gravity

Vf^2-Vo=2as -->>> Vf =sqrt(2(32.2)(260))= 129.0 ft/s

total velocity down and away from building at
sqrt(129^2+12^2) approx = 129.56 ft/s

can get the direction of the velocity by using trig with horizontal and vertical velocities

Answer 5

s = 1/2 at^2
t^2 = 2s/a = 2*260 / 32.2 = 16.15 =>
t = sqrt((16.15) = 4.019 sec

v = at = 32.2 * 4.019 = 129.4 f/sec

v(vector) = sqrt(129.4^2 + 12^2) = sqrt(16744 + 144)
= 129.95 f/s at arctan(-129.4 / 12) [magnitude at direction]

assuming horiz to right and down as "xbar = 12 and y = -129.4"