A toy car runs off the edge of a table that is 1.666 m high. The car lands 0.4049 m from the base of the table.

Question

1.How long does it take for the car to fall? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of s.

2.What is the horizontal velocity of the car? Answer in units of m/s.

Answer 1

When the car rolls of the table it has a zero vertical velocity and an unknown horizontal velocity V_x. So, this problem has to be analyzed in two parts: one dealing with the vertical dimension Y and the other, horizontal dimension  X.

Vertical dimension.

Since we are asked about the time that it will take the car to land on the ground  we should use the equation:

Y = Y_o + V_oyt + ½ gt²

The variables in this equation are:

Y      -       the final position of the car is zero it is on the ground)

Y_o     -       the initial position of the car above ground is 1.666 meters

V_oy    -       the initial vertical velocity of the car is zero (0)

a      -       the acceleration of gravity is -9.8 m/sec².

So plugging in the values we get:

0 = 1.666 + 0 + (1/2) (-9.8) t²

This then becomes

-.1.666 = - (1/2) (9.8) t²  or 1.666 = 4.9 t² 

So

t²  = 1.666/4.9 = 0.34 sec², and

t = 0.583 sec   

 

Horizontal dimension

 

The car is in the air for 0.583 seconds and it covers 0.4049 meters. Since

X = V_xt  we get 0.4049 = V_x(0.583)  and V_x = 0.752 m/sec