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

i wanna say it’s between b and d, leaning more towards b.

where's the following? you gotta add it to know what the choices are

explanation: