Question

A collision cart runs off the edge of a lab table that is 0.95 meters high and lands 0.40 meters from the base of the table. Calculate the speed of the collision cart when it leaves the table. Let g =10m/s²

Answer 1

Answer/Explanation:

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