Answer:
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;
[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43
[tex]t_{Moon}[/tex] ≈ 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.
