Question

How fast must a ball be rolled along the surface of a 70-cm high table so that when it rolls off the edge it will strike the floor at the same distance (70cm) from the point directly below the edge of the table?

Answer 1

change centimeters to meters and use -4.9 as the force of gravity to see how much time it takes to fall 70 cm

Answer 2

then you wil know that distance and the time it takes and that is a ratio of speed

Answer 3

-4.9t^2 + h = 0

when t^2 = h/4.9; so t = sqrt(h/4.9)

\[\frac{h}{\sqrt{h}/\sqrt{4.9}}\]
\[\frac{h\sqrt{4.9}}{\sqrt{h}}\]
\[\sqrt{4.9h}\] maybe?

Answer 4

70 (.01 m = 1 cm) = h

.7 m = 70 cm = h
\[\sqrt{4.9(.7)}\ mps\]
\[\sqrt{3.43}\ mps\]

Answer 5

i believe i might have a simple error in there someplace

Answer 6

nah, its right, it just looks odd when i do it another way

Answer 7

horizontal movement:
s = distance travelled = 0.70 m
let v = velocity as it leaves table
t = time to reasch floor

v = 0.70 / t

vertical movement
s = 0.70
u = initial vertical velocoty = 0
t = time taken
g = 9.8 m s-2

0.70 = 0t + 0.5*9.8 * t^2

t^2 = 0.70 / 4.9 = 1/7
t = 1 / sqrt7

plug this into equation for horizontal movement to get v

Answer 8

i'm stretching my memory of mechanics but i believe this is correct

Answer 9

lets see what amis comes up with

Answer 10

its correct

Answer 11

\[\frac{.7}{1/\sqrt{7}}=.7\sqrt{7}=\sqrt{3.43}=etc...\]