How long did it take the car to fall and how fast was the car going on the table?

Question

A toy car runs off the edge of table thats 1.225m high. If the car lands 0.400m from the base of the table.........

Answer 1

hey ,

Break the question down into two components ,horizontal and vertical . this is projectile motion so velocity is always constant in horizontal direction.

since it didn;t have any vertical velocity as it dropped v1y=0m/s

now you have gravitational acceleration of 9.8m/s^2 and the vertical height , 1.225m

so use motion formula

dy=v1y*t+.5at^2
1.225=.5(9.8)(t)^2
sqrt (1.225/.5(9.8))=t^2
t=0.5s.

as the time is constant in projectile motion for both horizontal and vertical components you could say:

vx = dx/t
vx=.4/.5
vx=0.8m/s speed with what the car was cruising on the table.

i hope it helped.

Answer 2

If the table is 1.225m high, then the time it took to drop is found by inverting
d = 1/2 a t^2 to get t = sqrt(2d/a)
or t = sqrt(2*1.225/9.8)

Given this, the horizontal speed of the car if it travels .4 meters in "t" seconds, is (and was as it left the table) .4/"t"

All you need now is a calculator!