Question

A toy car runs off the edge of a table that is 1.807 m high. The car lands 0.3012 m from the base of the table. 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. What is the horizontal velocity of the car? Answer in units of m/s.

Answer 1

The time it takes for the car to fall is 0.606 s and the horizontal velocity of the car is 0.497 m/s.

Step-by-step explanation:

To find the time it takes for the toy car to fall, we can use the kinematic equation:

h = (1/2)gt^2

Where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation, we can solve for t:

t = sqrt(2h / g)

Plugging in the values, we have:

t = sqrt(2 * 1.807 / 9.8) = 0.606 s

To find the horizontal velocity of the car, we can use the equation:

v = d / t

Where v is the velocity, d is the horizontal distance, and t is the time. Plugging in the values, we have:

v = 0.3012 / 0.606 = 0.497 m/s

Answer 2

a)
`t=0.60 s`

b)
`v_(ox) =0.5m/s`

Step-by-step explanation:

From the exercise we have initial height and final X position

`y_(o)=1.807m`

`X=0.3012m`

a) From the concept of free falling objects we have that

`y=y_(o)+v_(oy)t+(1)/(2)gt^(2)`

Since the car runs off the edge of the table, that means the car is moving in x direction with
`v_(oy)=0m/s` and at the end of the motion
`y=0m`

`0=1.807m-(1)/(2)(9.8)t^(2)`

Solving for t

t=± 0.6072 s

Since the time can not be negative, the answer is t=0.6072 s

b) To find the horizontal velocity of the car, we need to use the time that we just calculate

`X=v_(ox)t`

`v_(ox)=(X)/(t)=(0.3012m)/(0.6072s)  =0.5m/s`