Math Problem Statement

5. Sebuah roket diluncurkan hingga mencapai ketinggian maksimum 123 meter. Percepatan gravitasi di tempat peluncuran tersebut dianggap sebesar 10 m/s². Hitung kecepatan awal roket dan waktu yang diperlukan roket untuk mencapai ketinggian maksimum. 6. Sebuah benda dilempar vertikal ke atas dengan kecepatan awal 15 m/s. Jika benda tersebut mencapai ketinggian 30 meter dari posisi awalnya dan g = 10 m/s², tentukan lamanya benda tersebut di udara.

Solution

Let's solve the two problems from the image:


5. Sebuah roket diluncurkan hingga mencapai ketinggian maksimum 123 meter...

The rocket is launched vertically until it reaches a maximum height of 123 meters. The acceleration due to gravity gg is given as 10m/s210 \, \text{m/s}^2.

We need to calculate:

  • The initial velocity of the rocket
  • The time it takes for the rocket to reach maximum height.

1. Initial velocity:

We use the equation of motion: v2=u22ghv^2 = u^2 - 2gh where:

  • v=0m/sv = 0 \, \text{m/s} (velocity at maximum height),
  • h=123mh = 123 \, \text{m},
  • g=10m/s2g = 10 \, \text{m/s}^2 (gravitational acceleration),
  • uu is the initial velocity we need to find.

Rearranging the formula to solve for uu: u=2ghu = \sqrt{2gh} Substituting the known values: u=2×10×123=2460m/su = \sqrt{2 \times 10 \times 123} = \sqrt{2460} \, \text{m/s} Calculating: u49.6m/su \approx 49.6 \, \text{m/s} So, the initial velocity of the rocket is approximately 49.6m/s49.6 \, \text{m/s}.

2. Time to reach maximum height:

We use the formula: v=ugtv = u - gt Since v=0v = 0 at the maximum height, we can rearrange this to solve for time tt: t=ug=49.610=4.96secondst = \frac{u}{g} = \frac{49.6}{10} = 4.96 \, \text{seconds} Thus, the rocket takes approximately 4.96seconds4.96 \, \text{seconds} to reach its maximum height.


6. Sebuah benda dilempar vertikal dengan kecepatan awal 15 m/s...

An object is thrown vertically with an initial velocity of 15m/s15 \, \text{m/s}, and it reaches a height of 30meters30 \, \text{meters}. We need to determine how long the object stays in the air. g=10m/s2g = 10 \, \text{m/s}^2.

1. Time to reach maximum height:

At maximum height, the final velocity is zero. We use the following equation: v=ugtv = u - gt Rearranging to solve for tt: t=ug=1510=1.5secondst = \frac{u}{g} = \frac{15}{10} = 1.5 \, \text{seconds} So, it takes 1.5seconds1.5 \, \text{seconds} to reach maximum height.

2. Time to fall back to the ground:

The time to fall is the same as the time to rise, so the total time in the air is: ttotal=2×1.5=3secondst_{\text{total}} = 2 \times 1.5 = 3 \, \text{seconds} Thus, the object stays in the air for 3seconds3 \, \text{seconds}.


Would you like any more details or explanations for these solutions?

Here are 5 related questions to consider:

  1. How would the solution change if the acceleration due to gravity was different (e.g., 9.8 m/s²)?
  2. What is the relationship between initial velocity and maximum height in these types of problems?
  3. How can you calculate the total distance traveled by the rocket in problem 5?
  4. How does air resistance affect the motion of objects in vertical launch problems?
  5. How would the time of flight change in problem 6 if the initial velocity were doubled?

Tip: Always remember that the time to rise is equal to the time to fall when considering free fall under constant gravity without air resistance.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Vertical Motion under Gravity

Formulas

v^2 = u^2 - 2gh
v = u - gt
t = u/g

Theorems

Equations of motion under constant acceleration

Suitable Grade Level

Grades 10-12