Free fall motion – problems and solutions

1. A stone free fall from the height of 45 meters. If the acceleration due to gravity is 10 ms^{-2}, what is the speed of the stone when hits the ground?

__Known :__

Height (h) = 45 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ The final velocity of the stone when it hits the ground (v_{t})

__Solution :__

The equation of free fall motion :

v_{t}^{2} = 2 g h

The final velocity of the stone :

v_{t}^{2} = 2 (10)(45) = 900

v_{t} = √900 = 30 m/s^{2}

2. An object free fall from a height without the initial velocity. The object hits the ground 2 seconds later. Acceleration due to gravity is 10 ms^{-2}. Determine height

__Known :__

Time interval (t) = 2 seconds

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ Height (h)

__Solution :__

The equation of the free fall motion :

h = ½ g t^{2}

__Height :__

h = ½ (10)(2)^{2 }= (5)(4) = 20 meters

3.A 2-kg object free fall from a height of 20 meters above the ground. What is the time interval the object in air ? Acceleration due to gravity is 10 ms^{-2}

__Known :__

Height (h) = 20 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted ____:__ Time interval (t)

__Solution :__

The equation of free fall motion :

h = ½ g t^{2}

__Time interval :__

20 = ½ (10)(t^{2})

20 = (5)(t^{2})

20/5 = t^{2}

4 = t^{2}

t = √4

t = 2 seconds

4. Two objects, object 1 and object 2, are free fall from a height of h_{1} and h_{2} at the same time. If h_{1} : h_{2} = 2: 1, what is the ratio of the time interval of the object 1 to the object 2.

__Known :__

The height of the object 1 (h_{1}) = 2

The height of the object 2 (h_{2}) = 1

Acceleration due to gravity = g

__Wanted :__ t_{1} : t_{2}

__Solution :__

Object 1 :

h_{1} = 1/2 g t_{1}^{2}

2 = 1/2 g t_{1}^{2}

(2)(2) = g t_{1}^{2}

4 = g t_{1}^{2}

4/g = t_{1}^{2}

t_{1} = √4/g

Object 2 :

h_{2 }= 1/2 g t_{2}^{2}

1 = 1/2 g t_{2}^{2}

(2)(1) = g t_{2}^{2}

2 = g t_{2}^{2}

2/g = t_{2}^{2}

t_{2} = √2/g

The ratio of the time interval :

t_{1} : t_{2}

√4/g : √2/g

(√4/g)^{2} : (√2/g)^{2}

4/g : 2/g

4 : 2

2 : 1

5. An object dropped from a height of h above the ground. The final velocity when object hits the ground is 10 m/s. What is the time interval to reach ½ h above the ground. Acceleration due to gravity is 10 m/s^{2}.

__Known :__

The final velocity (v_{t}) = 10 m/s

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ The time interval to reach 1/2 h above the ground

__Solution :__

The height of h :

v_{t}^{2} = 2 g h

10^{2} = 2 (10) h

100 = 20 h

h = 100 / 20

h = 5 meters

The height of 1/2 h = 1/2 (5 meters) = 2.5 meters. The time interval needed to reach 2.5 meters above the ground :

h = 1/2 g t^{2}

2.5 = 1/2 (10) t^{2}

2.5 = 5 t^{2}

t^{2 }= 2.5 / 5 = 0.5 = (0.25)(2)

t = √(0.25)(2) = 0.5√2 = 1/2 √2 seconds

6.

The free fall motion of coconut (figure 1) and the motion of a ball thrown vertically upward to the highest point by a student (figure 2). Determine the kind of both motions.

Solution :

Figure 1 = __free fall motion__ = Acceleration

Figure 2 = __vertical motion__ = Deceleration

The correct answer is A.

7. A stone free fall from a building. The time interval needed by a stone to reach the ground is 3 seconds and acceleration due to gravity is 10 m/s^{2}. Determine the height of the building.

A. 15 m

B. 20 m

C. 30 m

D. 45 m

__Known :__

Time interval (t) = 3 seconds

Acceleration due to gravity (g) = 10 m.s^{-2}

__Wanted:__ Height of building (h)

__Solution :__

Known: time interval (t) and acceleration due to gravity (g), wanted: height (h) so use the equation of free fall motion: h = ½ g t^{2}

h = ½ (10)(3)

h = (5)(3)

h = 15 metes

The correct answer is A.

8. A fruit free fall from its tree at the height of 12 m above the ground. If acceleration due to gravity is g = 10 m/s^{2} and the friction of air ignored, then determine the height of the fruit above the ground after 1 second.

A. 7 m

B. 6 m

C. 5 m

D. 4 m

__Known :__

Height of tree (h) = 12 meters

Acceleration due to gravity (g) = 10 m/s^{2}

Time interval (t) = 1 second

__Wanted:__ The height of the fruit above the ground

__Solution :__

After 1 second, fruit free fall as far as :

h = ½ g t^{2} = ½ (10)(1)^{2} = (5)(1) = 5 meters

The height of the fruit above the ground after 1 second :

12 meters – 5 meters = 7 meters

The correct answer is A.

**What is free fall motion?****Answer**: Free fall motion refers to the motion of an object under the influence of gravitational force only, with no other forces (like air resistance) acting on it.**How does the acceleration due to gravity, often represented as $g$, affect a freely falling object?****Answer**: All objects in free fall near the surface of the Earth experience a constant acceleration due to gravity, $g$, which is approximately $9.81m/s$ downward. This means that the object’s velocity increases by this amount for each second of free fall.**If air resistance is negligible, how does the mass of an object influence its free fall acceleration?****Answer**: In the absence of air resistance, the mass of an object does not influence its free fall acceleration. All objects, regardless of their mass, will fall with the same acceleration due to gravity, $g$.**Why do astronauts appear to float inside the International Space Station (ISS) if gravity is still present there?****Answer**: Astronauts inside the ISS appear to float not because there’s no gravity, but because both the astronauts and the ISS are in a continuous state of free fall around the Earth. They’re essentially falling at the same rate as the ISS, creating a sensation of weightlessness.**What is the difference between weight and mass in the context of free fall?****Answer**: Mass is a measure of the amount of matter in an object and remains constant regardless of its location. Weight, on the other hand, is the force exerted on an object due to gravity. It varies depending on the gravitational field. During free fall, an object feels weightless because there’s no normal force acting on it, but its mass remains unchanged.**If an object is thrown upwards, what happens to its velocity as it rises? And what happens when it starts falling back down?****Answer**: When an object is thrown upwards, it decelerates under the influence of gravity. Its velocity decreases until it becomes zero at its highest point. As it starts falling back down, it accelerates due to gravity, increasing its velocity in the downward direction.**What is terminal velocity in the context of free fall?****Answer**: Terminal velocity is the constant maximum velocity reached by a falling object when the downward force of gravity is balanced by the upward force of air resistance. At this point, the object no longer accelerates and continues to fall at a constant speed.**How does the height from which an object falls influence the time it takes to reach the ground?****Answer**: The time it takes for an object to reach the ground is proportional to the square root of the height from which it falls (assuming no air resistance). An object dropped from a greater height will take longer to reach the ground than one dropped from a shorter height.**What happens to the potential energy of an object as it falls freely under gravity?****Answer**: As an object falls freely under gravity, its potential energy (relative to the ground) decreases. This decrease in potential energy is converted into kinetic energy, causing the object’s speed to increase.**If two objects of different shapes but the same mass are dropped from the same height in a vacuum, which will hit the ground first?**

**Answer**: In a vacuum, where there’s no air resistance, both objects will hit the ground at the same time. Their shape won’t matter because only gravity is acting on them, and their masses are the same.

You must be logged in to post a comment.