Question

A toy car runs off the edge of a table that is 1.225-m high. The car lands 0.400 m from the base of the table. How long did it take the car to fall?

Answer 1

using equation of motion
X=ut+[1/2] at^2
where
x =displacement = height 1.225
u = initial velocity = o
t = time
a = acceleration = gravitaional acceleration = 9.81m/s2
u=0 so
X = [1/2]at^2
(2x)/a=t^2
so
\[t=\sqrt{(2x)/a}\]
then just sub in x and a

Answer 2

Part b to the question :
how fast was the car going on the table ?

Answer 3

h = 1.225 - 0.4 = 0.825

v = sqrt(2 g h)
v = sqrt( 2 x 9.81 x 0.825)
v = sqrt(16.18)
v = 4.02 m/s

v = gt
t = v/g
t = 4.02/9.81
t = 0.4 s

Answer 4

@Liz13 v = 4.02 m/s