# Gerak Parabola Pre Test

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| By Dmorsilawati
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Dmorsilawati
Community Contributor
Quizzes Created: 2 | Total Attempts: 331
Questions: 5 | Attempts: 119  Settings  .

• 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

A. 0
Explanation
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.

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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.

• B.

• C.

2

• D.

1

• E.

0,5

C. 2
Explanation
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.

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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

D. Vo2 sin 2α / g
Explanation
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.

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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

B. 45 m
Explanation
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.

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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 Back to top