# 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.