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/s2
Wanted : The final velocity of the stone when it hits the ground (vt)
Solution :
The equation of free fall motion :
vt2 = 2 g h
The final velocity of the stone :
vt2 = 2 (10)(45) = 900
vt = √900 = 30 m/s2
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/s2
Wanted : Height (h)
Solution :
The equation of the free fall motion :
h = ½ g t2
Height :
h = ½ (10)(2)2 = (5)(4) = 20 meters
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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/s2
Wanted : Time interval (t)
Solution :
The equation of free fall motion :
h = ½ g t2
Time interval :
20 = ½ (10)(t2)
20 = (5)(t2)
20/5 = t2
4 = t2
t = √4
t = 2 seconds
4. Two objects, object 1 and object 2, are free fall from a height of h1 and h2 at the same time. If h1 : h2 = 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 (h1) = 2
The height of the object 2 (h2) = 1
Acceleration due to gravity = g
Wanted : t1 : t2
Solution :
Object 1 :
h1 = 1/2 g t12
2 = 1/2 g t12
(2)(2) = g t12
4 = g t12
4/g = t12
t1 = √4/g
Object 2 :
h2 = 1/2 g t22
1 = 1/2 g t22
(2)(1) = g t22
2 = g t22
2/g = t22
t2 = √2/g
The ratio of the time interval :
t1 : t2
√4/g : √2/g
(√4/g)2 : (√2/g)2
4/g : 2/g
4 : 2
2 : 1
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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/s2.
Known :
The final velocity (vt) = 10 m/s
Acceleration due to gravity (g) = 10 m/s2
Wanted : The time interval to reach 1/2 h above the ground
Solution :
The height of h :
vt2 = 2 g h
102 = 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 t2
2.5 = 1/2 (10) t2
2.5 = 5 t2
t2 = 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.
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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/s2. 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 t2
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/s2 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/s2
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 t2 = 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.
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