a. A remote-control car with a constant velocity drives off the top of a wall that is 10.0 m high and lands 4.60 m from the base of the wall. Draw a diagram of the problem. Label the known and unknown quantities. b. What is the car's speed before it drives off the top of the wall? c. What is the car's vertical speed just before impact? d. How far from the base of the wall would the remote-control car have landed if it had been subject to the gravitational field on the Moon (1.7 N/kg) ?

a. Please see the attached graph and drawing combined created with Microsoft Excel

b. Approximately 3.2217 m/s

c. Approximately 14.0 m/s  

d. Approximately 11.05 meters

Explanation:

a. The given parameters are;

The height of the wall, h = 10.0 m

The distance from the base of the wall the car lands = 4.60 m

The time, t, it takes the car to land is given by the equation for free fall as follows;

h = 1/2·g·t²

Where;

g = The acceleration due to gravity = 9.81 m/s²

From the equation for free fall, we have;

h/(1/2·g) = t²

∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278

The time it takes the car to land, t ≈ 1.4278 seconds

b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s

The horizontal speed of the car before it drives off the wall  ≈ 3.2217 m/s

c. The car's vertical speed just before impact is given by the following equation;

v = u + gt

Where;

u = The initial vertical speed = 0

t = The time it takes before impact ≈ 1.4278 seconds

∴ v = 9.81 × 1.4278 ≈ 14.0 m/s  

The car's vertical speed just before impact, v ≈ 14.0 m/s  

d. Whereby the car is subject to the gravitational field of the moon, we have;

Gravitational force per kilogram = 1.7 N/kg

∴ Gravitational acceleration = 1.7 m/s².

The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;

= √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43

≈ 3.43 seconds

The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s ×  3.43 seconds ≈ 11.05 meters

The horizontal distance covered, at the car's horizontal speed in the time of free fall  = The distance the car will land from the base of the wall ≈ 11.05 meters

The distance the car will land from the base of the wall ≈ 11.05 meters.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

prndulum needs thread, weight clock to time swings