Gerak Parabola Pre Test
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Questions and Answers
- 1.
Peluru ditembakkan dengan kecepatan awal 30 m/s dan sudut elevasi 30o. Pada saat mencapai titik tertinggi, kecepatannya adalah...
- A.
0
- B.
15 m/s
- C.
15√3 m/s
- D.
30 m/s
- E.
30√3 m/s
Correct Answer
A. 0Explanation
When a projectile reaches its highest point, its vertical velocity becomes zero. This is because at the highest point, the projectile momentarily stops moving upwards and starts moving downwards due to the force of gravity. Therefore, at the highest point, the velocity of the projectile is 0 m/s.Rate this question:
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- 2.
Bila besar sudut antara horizontal dan arah tembak suatu peluru adalah 45° , maka perbandingan antara jarak tembak dalam arah datar dan tinggi maksimum peluru adalah...
- A.
8
- B.
4
- C.
2
- D.
1
- E.
0,5
Correct Answer
C. 2Explanation
The correct answer is 2. When the angle between the horizontal and the firing direction of a bullet is 45 degrees, the ratio between the horizontal distance traveled by the bullet and its maximum height is 2. This can be explained using the concept of projectile motion. At this angle, the horizontal and vertical components of the bullet's initial velocity are equal. As a result, the time taken for the bullet to reach its maximum height is the same as the time taken for it to reach the same horizontal distance. Therefore, the ratio between the horizontal distance and the maximum height is 2.Rate this question:
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- 3.
Bola ditendang dengan sudut elevasi α dan kecepatan awal v0. Bila percepatan gravitasi bumi = g maka jarak terjauh yang dicapai oleh bola adalah….
- A.
Vo sin α / g
- B.
Vo sin α /2 g
- C.
Vo2 sin 2α /2 g
- D.
Vo2 sin 2α / g
- E.
Vo2 sin 2α /2 g
Correct Answer
D. Vo2 sin 2α / gExplanation
The correct answer is vo2 sin 2α / g. This formula represents the maximum range of a projectile launched at an angle α with an initial velocity v0, considering the acceleration due to gravity g. The term vo2 sin 2α represents the horizontal component of the initial velocity squared multiplied by the sine of twice the launch angle, which determines the range. Dividing this by g gives the maximum distance the ball can reach.Rate this question:
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- 4.
Sebuah peluru ditembakkan dengan kecepatan 60 m/s dan sudut elevasi 30°. Ketinggian maksimum yang dicapai adalah….
- A.
30 m
- B.
45 m
- C.
60 m
- D.
75 m
- E.
90 m
Correct Answer
B. 45 mExplanation
The maximum height reached by a projectile can be calculated using the formula: h = (v^2 * sin^2(θ))/(2g), where v is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity. Plugging in the given values (v = 60 m/s and θ = 30°) into the formula, we get h = (60^2 * sin^2(30°))/(2 * 9.8) = 45 m. Therefore, the correct answer is 45 m.Rate this question:
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- 5.
Sebuah bola ditendang dengan kecepatan awal 10 m/s. Jika sudut elevasinya 45o, dan percepatan gravitasi bumi 10 m/s2, maka jarak terjauh yang dicapai bola adalah…
- A.
5 m
- B.
10 m
- C.
15 m
- D.
20 m
- E.
40 m
Correct Answer
B. 10 mExplanation
When a ball is kicked with an initial velocity of 10 m/s and an elevation angle of 45 degrees, the maximum distance it can reach can be calculated using the range formula. The range formula is given by R = (v^2 * sin(2θ))/g, where R is the range, v is the initial velocity, θ is the elevation angle, and g is the acceleration due to gravity. Plugging in the given values, we get R = (10^2 * sin(90))/10 = 10 m. Therefore, the ball will reach a maximum distance of 10 meters.Rate this question:
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Jul 22, 2024Quiz Edited by
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Sep 18, 2019Quiz Created by
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