Question

A collision 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?

Answer 1

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