Acollision cart runs off the edge of a lab table that is .95 meters high. how fast did the collision cart leave the tabletop if the cart lands .40 meters from the base of the table?

From the given values we can generate the equations:

x = vx*t 
y = 0.95 - 1/2*g*t^2 

where x refers to displacement in horizontal direction, y refers to displacement in vertical direction, and t is time, while vx is velocity of the cart in the x direction, and g is the acceleration due to gravity = 9.8 m/s^2

At it's landing point we have x = 0.4, y = 0: 

0.4 = vx*t 
0 = 0.95 - 1/2*9.8*t^2 

Therefore combining the equations and solving:
t = 0.44 s s 
vx = 0.91 m/s

report!

explanation:

physics means: "the branch of science concerned with the nature and properties of matter and energy. the subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms."

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y